Chapter 13

Haley Norris

13 Chapter 13

Chapter 3 Curve Sketching and Optimization

3.3 Absolute Extrema

Until now, we have only discussed one type of extrema: local extrema. Recall that a local extremum occurs only in a neighborhood (or localized area) around it. Another way to think of this is that the function must exist on both sides of a local extremum. However, a function need not be continuous at a local extremum (we will explore this more soon!).

In this section, we will learn how to find another type of extrema: absolute extrema. Absolute extrema are the absolute maximum and absolute minimum (values) of a function. You can think of the word “absolute” as meaning exactly as it sounds: Absolute extrema are the absolute highest (maximum) and lowest (minimum) values of a function.

Figure 3.3.1 below shows the graph of a function f that is defined on the interval [-10, 8] and has an absolute maximum of 5 occurring at x = 8 and an absolute minimum of -2 occurring at X = -3:

Figure 3.3.1: Graph of a function f that is defined on the interval [-10,8]

Notice f has local maxima at x = -8 and x = 2 and local minima at x = -3 and x = 5; local and absolute extrema can occur simultaneously!

Learning Objectives:

In this section, you will learn how to identify absolute extrema graphically as well as how to use calculus to find the absolute extrema of a function on a closed interval on which the function is continuous. Upon completion you will be able to:

Identify any local and absolute extrema of a function by examining the graph of the function.
State the Extreme Value Theorem.
Identify any absolute extrema of a function on a given interval (open or closed) by examining the graph of the function.
Describe the conditions in which the Closed Interval Method can be used to calculate the absolute extrema of a function on a given interval.
Perform the Closed Interval Method to determine the absolute maximum and minimum values of a function on a closed interval on which the function is continuous.
Determine the absolute maximum and minimum values of a function on an interval in which the Closed Interval Method cannot be performed by using alternative strategies, such as graphing the function.

Identifying Absolute Extrema Graphically

As stated previously, the absolute maximum and absolute minimum of a function are the highest (largest) and lowest (smallest) y-values of the function. Together, the absolute maximum and minimum are called the absolute extrema of a function. We refer to either an absolute maximum or absolute minimum as an absolute extremum.

We now formally define absolute extrema:

Definition 3.10

f (c) is an absolute maximum of f if f (c) ≥ f (x) for all x in the domain of f.
f (c) is an absolute minimum of f if f (c) ≤ f (x) for all x in the domain of f.
f (c) is an absolute extremum of f if f (c) is an absolute maximum or minimum.

Absolute extrema can occur simultaneously with local extrema inside a function or at an endpoint, as seen in Figure 3.3.1. Because local extrema must occur in a neighborhood of a function, they cannot occur at endpoints (remember, the function must exist on both sides of a local extremum, even if the function is not continuous at the local extremum). However, absolute extrema can occur anywhere!

In contrast to local extrema, a function can have only one absolute maximum and one absolute minimum (y-values). Although, the absolute maximum and minimum can occur at infinitely many x-values. This type of behavior can be seen in the function defined on [-9, 9] shown in Figure 3.4.2. There is only one absolute maximum, y = 5, and one absolute minimum, y = l, but they both occur at multiplex-values.

Figure 3.3.2: Graph of a function f that has absolute extrema occurring at multiple x-values

The following table summarizes the similarities and differences of local and absolute extrema:

Local Extrema Absolute Extrema

Occur inside the function Occur inside the function at

local extrema or at an endpoint

Multiple local extrema (y-values) At most one absolute max. and one absolute

min. (y-values)

Before we get into the details of using calculus to find the absolute extrema of a function, let’s practice identifying absolute extrema and review finding local extrema graphically!

Example 1 Given the graph off shown in Figure 3.4.3, find any local and absolute extrema off as well as where they occur.

Figure 3.3.3: Graph of a continuous function f

Solution: First, let’s start by finding any absolute extrema. To find the absolute maximum, we look for the highest (largest) y-value on the graph. The absolute maximum is 3, and it occurs at x = 4. To find the absolute minimum, we look for the lowest (smallest)

y-value on the graph. Because j(x) —-+ -oo as x —-+ -oo and x —-+ oo, the function continues toward negative infinity on both ends and never reaches a lowest y-value (1•.e., 1·t never “st .ops”).

In terms of local extrema, f also has a local maximum of 3 at x = 4. Recall that local and absolute extrema can occur simultaneously. However, the function does not have a local minimum.

In summary, the absolute maximum of f is 3, and it occurs at x = 4. There is no absolute minimum. f also has a local maximum of 3 occurring at x = 4, but there is no local minimum.

Try It 1

Given the graph of f shown in Figure 3.4.4, find any local and absolute extrema of f as well as where they occur.

Figure 3.3.4: Graph of a continuous function f

Example 2 Given the graph off shown in Figure 3.4.5, find any local and absolute extrema off

as well as where they occur.

Figure 3.3.5: Graph of a discontinuous function f

Solution: Let’s start with finding the absolute extrema. We see .f does not have an absolute maximum nor an absolute minimum because of the behavior of the function near the

vertical asymptote, x = 0. Specifically, because .f (x) —+ -oo as x —+ o-, the function continues to decrease and never reaches a lowest y-value. Thus, there is no absolute mm1mum.

Likewise, because f (x) —+ oo as x —+ o+, the function continues to increase and never reaches a highest y-value. Thus, there is no absolute maximum. Note that there is also no absolute maximum because of the end behavior of the function (both ends tend to positive infinity).

Looking for local extrema, we see f has local minima of -4 at x = -12 and -8 at x = 6. While we might be tempted to say f has a local maximum at x = -6, thefunction is undefined at x = -6, so there cannot be any type of extrema.

In summary, f does not have any absolute extrema. .f has local minima of -4 and -8 occurring at x = -12 and x = 6, respectively, but it does not have any local maxima.

Example 3 Given the graph off shown in Figure 3.4.6, find any local and absolute extrema off as well as where they occur.

Figure 3.3.6: Graph of a discontinuous function f

Solution: First, let’s look for absolute extrema, if they exist. The absolute highest y-value of the function is 6, and it occurs at x = -2. Again, we may be tempted to say there is an absolute minimum at x = 2, but the function is undefined at x = 2, so there cannot be any type of extrema. We can also think of the function as never “reaching” the point at x = 2 (hence, the hole in the graph). Therefore, the function has no absolute minimum.

In terms of local extrema, f has a local maximum of 5 at x = 2. It may seem surprising that there is a local maximum at the point (2, 5). However, remember that a function does not have to be continuous at a local extremum; the function just has to exist on both sides of the local extremum (i.e., a local extremum must occur in a neighborhood

of the function). In this case, the 71-value 5 is greater than the function values on either side of x = 2, so 5 is a local maximum, even though the function is not continuous at

X = 2.

f does not have any local minima. There is no local minimum at x = 6 because local extrema cannot occur at endpoints (local extrema must occur in neighborhoods in which there is function on both sides of the extrema). There is no local minimum at x = 2 because, again, f is undefined at x = 2.

In summary, f has an absolute maximum of 6 occurring at x = -2, but it does not have an absolute minimum. f has a local maximum of 5 occurring at x = 2, but it does not have any local minima.

Try It 2

Given the graph of f shown in Figure 3.4.7, find any local and absolute extrema of f as well as where they occur.

Figure 3.3.7: Graph of a discontinuous function f

Absolute Extrema on an Interval

In the previous examples, we were looking for the extrema of a function on its entire domain. We will now learn how to find absolute extrema on a specified interval (either open, half open, or closed). If the interval given is open, the corresponding endpoints (x-values) are not included and the interval will have parentheses around both endpoints. If the interval is half-open, the endpoint with a parenthesis is not included, but the endpoint with a bracket is included. If the interval is closed, then both endpoints are included and both will have brackets. There is a very important theorem called the Extreme Value Theorem which states

that if a function is continuous on a closed interval [a, b], then the function must have both an absolute maximum and absolute minimum on the interval:

Theorem 5 Extreme Value Theorem

If a function is continuous on a closed interval [a,b], then it must have both an absolute maximum and an absolute minimum on the interval.

Example 4 Given the graph off shown in Figure 3.4.8, find the absolute maximum and minimum off on each of the following intervals.

Figure 3.3.8: Graph of a continuous function f

(a) [O, 2]

(b) [-2,1]

Solution: (a) First, note that [O, 2] is a closed interval and that the function f is continuous on the interval (in fact, it is continuous everywhere!). Thus, the Extreme Value The orem says the function must have an absolute maximum and absolute minimum. Only looking at and considering the graph of the function from x = 0 to x = 2

(including these endpoints), we see the absolute maximum is 6, and it occurs at

x = 2. The absolute minimum is 2, and it occurs at x = l.

(b) Again, the interval [-2, 1] is closed, and we know f is continuous on the interval. Only looking at and considering the graph of the function from x = – 2 to x = l

(including these endpoints), we see the absolute maximum is 6, and it occurs at x = -1. The absolute minimum is 2, and it occurs at x = -2 and x = l. Remember, a function can have only one absolute maximum and one absolute

minimum (y-values), but they can occur at multiplex-values.

Try It 3

Given the graph of f shown in Figure 3.4.9, find the absolute maximum and minimum of f on each of the following intervals.

Figure 3.4.9: Graph of a continuous function f

(a) [-3,1]

(b) [-2, 4]

Now, let’s practice working examples in which we find where absolute extrema occur on a variety of types of intervals! Remember, the Extreme Value Theorem only guarantees there are absolute extrema if the function is continuous on a closed interval.

Example 5 Given the graph of .f shown in Figure 3.3.10, determine where any absolute extrema occur on each of the following intervals.

Figure 3.3.10: Graph of a continuous function .f

(a) (-2, 3)

(b) [-1, 2)

(c) (-00,00)

(d) [0,4]

Solution: (a) Notice first that the interval we are given, (-2, 3), is an open interval (parentheses around both endpoints). This means that we do not include the endpoints in our consideration for absolute extrema. It also means the Extreme Value Theorem does not apply because the interval is not closed.

To just focus on the part of the graph on the interval (-2, 3) and remember the endpoints are not included, it may be helpful to mark (using your pencil) an open circle, 0, on the graph at the points at x = -2 and x = 3. This may help you remember not to consider these points as absolute extrema. Remember, a hole on the graph of a function means the function never “reaches” the point. See Figure

3.3.11.

Figure 3.3.11: Graph off on the interval (-2, 3)

Looking only at the interval (-2, 3), we see f has an absolute maximum at x = 2. Although there would be an absolute minimum at x = -2 if the endpoint was included, there is no absolute minimum because the interval is open so the point at x = -2 is not included. Again, we can think of the function as never “reaching” the point at x = -2 because it is not included (as indicated by the hole at the point if you drew an open circle on the graph).

In summary, on the interval (-2, 3), f has an absolute maximum at x = 2, but it does not have an absolute minimum. Because the question asked us to find “where” any absolute extrema occur, we only need to identify the x-value(s) of the absolute extrema.

(b) The interval [-1, 2) is said to be half-open (or half-closed). This means the point at x = – l is included and can be considered when finding absolute extrema, but the point at x = 2 is not included and cannot be considered.

Using the technique above, we will draw a solid dot, •, on the graph at x = -l and an open circle, Q, on the graph at x = 2 to help us remember which endpoint is included and which is not. See Figure 3.3.12.

Figure 3.3.12: Graph off on the interval [-1, 2)

Looking at the interval [-1, 2), we see f does not have an absolute maximum because the function never “reaches” the point at x = 2 (as indicated by the hole at the point if you drew an open circle on the graph). The smallest y-value on the interval occurs at x = – l. As indicated by the solid dot on the graph, this endpoint is included. Thus, the function has an absolute minimum at x = -1.

(c) The interval (-oo, oo) is considered an open interval; it is the real number line. So we must find the absolute extrema of the entire function (not just on a finite interval). Looking at the entire graph, we see there are no absolute extrema because of the end behavior of the function. As x —-+ -oo, f (x) —-+ oo, so there is no absolute maximum. As x—-+ oo, f(x) —-+ -oo, so there is no absolute minimum.

(d) The interval [O, 4] is closed because there are brackets around both endpoints. This means the points at both x = 0 and x = 4 are included, and we can consider them both when determining where the absolute extrema occur. To help us focus on this part of the graph and remember both endpoints are included, we will draw solid dots, •, on the graph at both endpoints. See Figure 3.3.13.

Figure 3.3.13: Graph off on the interval [O, 4]

We see that f has an absolute maximum at x = 2 and an absolute minimum at x = 4 on the interval [O, 4]. Note that because this interval is closed, and f is continuous on the interval, the Extreme Value Theorem guarantees the function will have both an absolute maximum and absolute minimum.

Example 6 Given the graph of .f shown in Figure 3.3.14, determine where any absolute extrema occur on each of the following intervals.

Figure 3.3.14: Graph of a discontinuous function .f

(a) [-16,-6)

(b) (2,4)

(c) [-6,6]

Solution: (a) The interval [-16, -6) is half-open (or half-closed). This means that the point at x = -16 is included and can be considered when finding the absolute extrema, but the point at x = -6 is not included and cannot be considered.

Using the technique from the previous example, we will draw a solid dot, •, on the graph at x = -16 and an open circle, 0, on the graph at x = -6 to help us remember which endpoint is included and which is not. See Figure

Figure 3.3.15: Graph off on the interval [-16, 6)

Looking at the interval [-16, -6), we see f does not have an absolute maximum because it never “reaches” the point at x = -6 (as indicated by the hole at the point if you drew an open circle on the graph). The smallest y-value on the interval occurs at x = -12. Thus, the function has an absolute minimum at

X = -12.

(b) The interval (2, 4) is open, so neither endpoint is included. To help us focus on this part of the graph, we will draw open circles, 0, at the points at both x = 2 and x = 4. See Figure 3.3.16.

Figure 3.3.16: Graph off on the interval (2, 4)

We see that .f does not have any absolute extrema on the interval (2, 4). The function never “reaches” either of the endpoints (as indicated by the holes at the endpoints), so it does not have an absolute maximum nor an absolute minimum.

(c) Although the interval [-6, 6] is closed, the function is not continuous on the interval, so it is not guaranteed to have absolute extrema.

To help us focus on this part of the graph and remember that both endpoints are included, we will draw solid dots, •, on the graph at both endpoints. See Figure 3.3.17.

Figure 3.3.17: Graph of .f on the interval [-6, 6]

f does not have absolute extrema on the interval [-6, 6] due to the behavior of the function near the vertical asymptote, x = 0. Specifically, because f(x)–+ -oo as x –+ o-, there is no absolute minimum. Likewise, because f (x) –+ oo as x –+ o+, there is no absolute maximum.

Try It 4

Given the graph of f shown in Figure 3.3.18, determine where any absolute extrema occur on each of the following intervals.

Figure 3.3.18: Graph of a discontinuous function f

(a) [-5, 2)

(b) (2, 7]

(c) (-3, 5]

Determining Absolute Extrema Using Calculus

Now that we have learned how to find absolute extrema graphically, let’s learn how to find the absolute extrema of a function on an interval when we are only given the rule of the function (and not the graph of the function). The Extreme Value Theorem guarantees that as long as a function is continuous on a closed interval, it will have an absolute maximum and absolute minimum on the interval. So we will start by learning how to find the absolute extrema of a function on a closed interval on which the function is continuous.

To establish the calculus method for finding the absolute extrema of a function on a closed interval on which the function is continuous, we will use an important fact established previously: Absolute extrema either occur simultaneously with local extrema inside a function or at an endpoint. Recall that local extrema occur at the critical values of a function. Thus, as a result, absolute extrema will either occur at the critical values of a function or at an endpoint:

Theorem 6

If a function has an absolute extremum, it must occur at a critical value or at an endpoint.

To find the absolute extrema of a function on a closed interval on which the function is continuous, we need a test that compares the values of the function at the critical values in the interval and the endpoints of the interval. This test is called the Closed Interval Method. It consists of three steps:

Finding Absolute Extrema Using the Closed Interval Method

Note: Later in this section, we will use technology to help us find absolute extrema if we cannot use the Closed Interval Method because the interval is not closed or the function is not continuous on the interval, or both.

Example 7 Find the absolute maximum and minimum of the function .f(x) = x3 – 2×2 – 4x + 6 on each of the following intervals.

(a) [-2, 3]

(b) [O, 4]

(c) [-3, -1]

Solution: (a) Because the interval [-2, 3] is closed (as indicated by the brackets around both endpoints), we can attempt to use the Closed Interval Method to find the absolute extrema:

1. Determine the domain of f, and check that the function is continuous on the closed interval:

.f is a polynomial, so its domain is (-oo, oo). Hence, the function is continu ous on (-oo, oo); recall from section 1.4 that polynomials, rational functions, power functions, exponential functions, logarithmic functions, and combina tions of these are continuous on their domain. Thus, the function is continu ous on the closed interval [-2, 3], so we can proceed with the Closed Interval Method.

2. Find the critical values of .f:

Recall that the critical values off are the x-values where f'(x) = 0 or f'(x) does not exist, and they are in the domain off. To find the critical values of f, we must first find f’ (x):

f ( x) = x3 – 2×2 – 4x + 6 ==}

J’ ( x) = 3×2 – 4x – 4

Next, we find the x-values where f'(x) = 0 or J'(x) does not exist. Because f’ is a polynomial and has a domain of all real numbers, it will exist everywhere. Thus, we only need to find the x-values where f'(x) = 0:

J'(x) = 0 ==}

3×2 – 4x – 4 = 0

(3x + 2)(x – 2) = 0

2

==} x = – – and x = 2

3

Because the domain of f is all real numbers, x = – i and x = 2 are both in the domain. Thus, both are critical values of f.

3. Evaluate f at the critical values in the interval and the endpoints of the interval:

Because both critical values (x = -i and x = 2) are in the interval [-2, 3], we will find the value of the function at each critical value and each endpoint of the interval (x = -2 and x = 3):

f (x) = x3 – 2×2 – 4x + 6 ==}

f ( – -2)

= ( – -2)

3

– 2 (

– -2)

2

– 4 ( – -2)

+ 6 = -202

7.4815

3 3 3 3 27

f (2) = (2)3 – 2(2)2 – 4(2) + 6 = -2

f(-2) = (-2)3 – 2(-2)2 – 4(-2) + 6 = -2

f (3) = (3)3 – 2(3)2 – 4(3) + 6 = 3

Looking at the function values, we see the largest is 2 °; at x = . The smallest function value is -2, and it occurs at both x = -2 and x = 2. Thus, on the interval [-2, 3], the absolute maximum of f is 2 °;,and the absolute

minimum off is -2.

We can check our work by looking at the graph off shown in Figure 3.3.19, but using the Closed Interval Method allows us to find the absolute extrema completely algebraically in order to get exact answers.

Figure 3.3.19: Graph of the function f(x) = x3 – 2×2 – 4x + 6

(b) Because the interval we are given, [0,4], is closed and we already know the function j(x) = x3 – 2×2 – 4x + 6 is continuous everywhere, the Closed Interval Method applies.

We found the critical values off in part (a) (x = -i and x = 2), so we can go straight to step 3 of the Closed Interval Method:

3. Evaluate f at the critical values in the interval and the endpoints of the interval:

Because only one of the critical values, x = 2, is in the interval [0, 4], we will only find the value of the function at this critical value as well as each endpoint of the interval:

f (x) = x3 – 2×2 – 4x + 6 ==}

j (2) = (2)3 – 2(2)2 – 4(2) + 6 = -2

f (0) = (0)3 – 2(0)2 – 4(0) + 6 = 6

f(4) = (4)3 – 2(4)2 – 4(4) + 6 = 22

Looking at the function values, we see the largest is 22 at x = 4. The smallest function value is -2, and it occurs at x = 2. Thus, on the interval [0, 4], the absolute maximum of f is 22, and the absolute minimum of f is -2.

We can verify our answer by looking at the graph off again in Figure 3.3.19.

(c) Again, because we are given a closed interval, [-3, -1], andwe already know the function f(x) = x3 – 2×2 – 4x + 6 is continuous everywhere, the Closed Interval Method applies.

We found the critical values off in part (a) (x = -i and x = 2), so we can go straight to step 3 of the Closed Interval Method again:

3. Evaluate f at the critical values in the interval and the endpoints of the interval:

Neither of the critical values is in the interval [-3, -1], so we will only find the value of the function at each endpoint of the interval:

f (x) = x3 – 2×2 – 4x + 6 ==}

f(-3) = (-3)3 – 2(-3)2 – 4(-3) + 6 = -27

f(-1) = (-1)3-2(-1)2 -4(-1) +6 = 7

Looking at the function values, we see the largest is 7 at x = – l. The smallest function value is -27, and it occurs at x = -3. Thus, on the interval [-3, -1], the absolute maximum off is 7, and the absolute minimum off is -27.

We can verify our answer by looking at the graph off again in Figure 3.3.19.

Try It 5

Find the absolute maximum and minimum of the function f(x) = 1/4x^4 – 9/2 x^2 + 6 on each of the following intervals.

(a) [-5, 4]

(b) [1, 6]

Example 8 Find the absolute maximum and minimum of the function f(x) = 2x: 10 on each of

the following intervals.

(a) [4,8]

(b) [-5,0]

Solution: (a) Because the interval [4, 8] is closed, we can attempt to use the Closed Interval Method to find the absolute extrema:

1. Determine the domain of f, and check that the function is continuous on the closed interval:

f(x) = 2: l0 is a rational function, so the denominator must be nonzero.

There are no other domain issues (there is no even root or logarithm). Thus, we just need to ensure x – 3 -/- 0, which gives x -/- 3. Therefore, the domain off is (-oo,3) U (3,oo), so f is continuous on (-oo,3) U (3,oo).

Because [4, 8] is a closed interval and f is continuous on this interval (note that x = 3 is not in the interval), we can proceed with the Closed Interval Method.

2. Find the critical values off:

Thecritical values off are the x-values where f'(x) = 0 or f'(x) does not exist, and they are in the domain off. To find the critical values off, we

must first find f’ (x) using the Quotient Rule:

f(x)= 2×2-_10 ==}

x-3

f'(x)= (x- 3) (d (2x – 10)) – (2x

(x – 3)2

(x – 3)(4x) – (2×2 – 10) (1)

(x – 3)2

4×2 – 12x – 2×2 + 10

(x – 3)2

2×2 – 12x + 10

(x – 3)2

-10) (d (x- 3))

Now, we find the x-values where f'(x) =0 or f'(x) does not exist. f'(x) =0 when the numerator equals 0, and f'(x) does not exist when the denominator equals zero (remember we are looking at the domain off’ when determining where it does not exist). First, let’s find the x-values where f'(x) =0:

J'(x) = 0 ==}

2×2 – 12x + 10 = 0

2 (x2 – 6x + 5) = 0

2(x – l)(x – 5) = 0

==} x = l and x = 5

Next, to find the x-values where f'(x) does not exist, we set the denominator equal to zero:

J'(x)DNE ==}

(x – 3)2 = 0

((x – 3)2)21

= (0)2

x-3=0

x=3

Because the domain off is (-oo, 3) U (3, oo), only x = l and x = 5 are in the domain. Thus, these are the only critical values off.

3. Evaluate f at the critical values in the interval and the endpoints of the interval:

Because only one critical value, x = 5, is in the interval [4, 8], we will only find the value of the function at this critical value as well as each endpoint of

the interval:

f(x)= 2×2 -_10 ==;,

x-J

f (5) = 2(5; 10 = 20

2 10

f(4) = (1 = 22

!(8)= 2(8)2- 10= 118= 23.6

8-3 5

Looking at the function values, we see the largest is 1;8 at x = 8. The smallest function value is 20, and it occurs at x = 5. Thus, on the interval [4, 8], the

absolute maximum off is 1;8,

and the absolute minimum off is 20.

We can verify our answer by looking at the graph off shown in Figure 3.4.20.

Figure 3.3.20: Graph of the function j(x) = 2 2 10

(b) Because the interval [-5, 0] is closed and we already know f (x) = 2:2

10 is continuous on (-oo, 3) U (3, oo), the Closed Interval Method applies because the function is continuous on the interval [-5, 0].

We found the critical values of f in part a (x = 1 and x = 5), so we can go straight to step 3 of the Closed Interval Method:

3. Evaluate f at the critical values in the interval and the endpoints of the interval:

Neither of the critical values is in the interval [-5, 0], so we will only find the

value of the function at each endpoint of the interval:

.( ) 2×2 10

j X =

–

x-3

===}

f(-5)= 2(-5)2 10= -5

-5-3

2(0)2 – 10 10

f(O) = = – 3.3333

0-3 3

Looking at the function values, we see the largest is 1 °at x = 0. The smallest function value is -5, and it occurs at x = -5. Thus, on the interval [-5, 0], the absolute maximum off is \0, and the absolute minimum off is -5.

We can verify our answer by looking at the graph off again in Figure 3.4.20.

Try It 6

Find the absolute maximum and minimum of the function f(x)=x^3/x^2 – 1 on each of the following intervals.

(a) [-0.5, 0.5]

(b) [4/3, 6]

What if we cannot use the Closed Interval Method?

In the previous examples, the function given was continuous on a closed interval. What do we do if the function is not continuous on the interval or the interval is not closed, or both? Although there are calculus techniques we can use to find the absolute extrema in such cases, for the purposes of this textbook, we will calculate the critical values and then graph the function using technology to determine if there are any absolute extrema.

We must also remember to consider the information about the endpoints of the interval when we look at the graph of the function on our calculator. If the interval is open, that means the endpoints are not included, which could affect the answer in some situations.

Example 9 Find the absolute maximum and minimum of the function f(x) = :2 ;:; on each of the following intervals, if they exist.

(a) [-10, 3]

(b) (-2,4)

(c) (-13,-10]

Solution: (a) Because the interval [-10, 3] is closed, we can attempt to use the Closed Interval Method to find any absolute extrema:

1. Determine the domain of f, and check that the function is continuous on the closed interval:

f is a rational function, so the denominator must be nonzero. There are no other domain issues (there is no even root or logarithm). Thus, we just need to ensure x + 4 -/:- 0, which gives x -/:- -4. Therefore, the domain of f is

(-oo, -4) U (-4, oo), so f is continuous on (-oo, -4) U (-4, oo).

Even though [-10, 3] is a closed interval, f is not continuous on this interval because x = -4 is in the interval. Therefore, we cannot proceed with the Closed Interval Method.

Before attempting to find any absolute extrema by graphing the function on our calculator, we will calculate the critical values of f. Knowing the critical values will help us adjust our viewing window so we can see the pertinent information on the graph, as well as help us verify the exact values of potential x-coordinates where the extrema occur.

The critical values off are the x-values where .f'(x) = 0 or .f'(x) does not exist, and they are in the domain off. To find the critical value sof f, we will start by finding f’ (x) using the Quotient Rule:

f(x)= x2 +9 ==;,

x+4

I (x=) (X + 4) ( d ( x2 + 9)) – (x2 + 9) ( d ( X + 4))

f (x+4)2

(x + 4)(2x) – (x2 + 9) (1)

(x + 4)2

2×2 + 8x – x2 – 9

(x + 4)2

x2 + 8x – 9

(x + 4)2

Now, we find the x-values where .f'(x) = 0 or .f'(x) does not exist . .f'(x) = 0 when the numerator equals 0, and f'(x) does not exist when the denominator equals zero (remember we are looking at the domain of f’ when determining where it does not exist). First, we will find the x-values where f'(x) = 0:

J'(x) = 0 ==;,

x2 + 8x – 9 = 0

(x + 9)(x – 1) = 0

==;, x = -9 and x = l

Next, to find the x-values where f'(x) does not exist, we set the denominator

equal to zero:

J'(x) DNE ==}

(x + 4)2 = 0

((x + 4)2) 2 = (0)2

x+4=0

X = -4

Because the domain off is (-oo, -4) U (4, oo), only x = -9 and x = l are in the domain. Thus, these are the only critical values off.

Using this information, we can adjust our calculator’s viewing window to be sure we can see the behavior of the function on the interval [-10, 3], as well as at the critical values. See Figure 3.4.21.

Figure 3.3.21: Graph of the function f(x) = :1

Regardless of whether or not the endpoints of the interval [-10,3] are included in our consideration (which they both are), the behavior of the function near the vertical asymptote at x = -4 shows there is no absolute maximum nor absolute minimum on the interval.

Specifically, because f (x) –+ -oo as x –+ -4-, there is no absolute minimum. Likewise, because f(x) –+ oo as x–+ -4+, there is no absolute maximum.

(b) Because the interval (-2, 4) is open, we know immediately that we cannot use the Closed Interval Method (regardless of whether or not the function is continuous on the interval, which it is).

Again, we will look at the graph off shown in Figure 3.4.21 to locate any absolute extrema on the interval (-2, 4). When looking at the graph on the interval (-2, 4), we must remember that the endpoints are not included because the interval is open. Recall that you can draw open circles at each endpoint on your own paper to help remind you that the endpoints are not included.

Looking at the graph, we see that the function never “reaches” an absolute max imum at x = -2 because that endpoint is not included. However, it appears as though the function has an absolute minimum near x = l. Because we know from part a the function has a critical value at x = l, we can say with confidence that there is an absolute minimum at x = l.

Even though we could use the First Derivative Test or the Second Derivative Test

to verify there is a minimum at x = l, this is not necessary because we can see from the graph there is a minimum. Knowing the critical values of f helps us determine the exact location of the absolute minimum. Now, we can find the absolute minimum exactly by evaluating f at x = 1:

f ( X)= X2+ 9

x+4

!(1)=( 1)2+9=2

1+4

Thus, on the interval (-2, 4), the absolute minimum of f) is 2, but there is no absolute maximum.

(c) We are given a half-open (or half-closed) interval, (-13, -10], so we know imme diately we cannot use the Closed Interval Method because we do not have a closed interval (regardless of whether or not the function is continuous on the interval, which it is).

Again, we look at the graph off shown in Figure 3.4.21 to locate any absolute extrema on the interval (-13, -10]. When looking at the graph on the interval (-13, -10], we must remember that the endpoint x = -13 is not included, but the endpoint x = -10 is included. Remember, you can draw an open circle at the point at x = -13 and a solid dot at the point at x = -10 on your own paper to help remind you which endpoint is included and which is not.

Looking at the graph, we see that the absolute maximum on the interval (-13, -10] occurs at x = -10, but the function never “reaches” an absolute minimum at x = -13 because that endpoint is not included. We can find the absolute maxi mum exactly by evaluating f at x = -10:

f(x)= x2 +9

x+4

f(-10) = (-l0)2+ 9 = 109 -18.1667

-10 + 4 -6

Thus, on the interval (-13, -10], the absolute maximum of f is – 1 9, butthere is no absolute minimum.

Example 10 Find the absolute maximum and minimum of the function f(x) =(:; 2 on the interval (-4, 15), if they exist.

Solution: The interval (-4, 15) is open, so we know immediately we cannot use the Closed Interval Method (regardless of whether or not the function is continuous on the interval, which it is not).

Before attempting to find any absolute extrema by graphing the function on our cal culator, we will calculate the critical values of f so we know exactly where there may be absolute extrema and can adjust the viewing window on our calculator accordingly.

The critical values off are the x-values where .f'(x) = 0 or J'(x) does not exist, and they are in the domain off. To find the critical values off, we will start by finding f’ (x) using the Quotient Rule:

‘(x)= (x + 2)2 C (x – 5)) – (x 5) (d ((x + 2)2))

f ((x + 2)2)2

(x + 2)2(1) – (x – 5) (2(x + 2) (d (x + 2)))

(x + 2)4

(x + 2)2 – (x – 5)(2(x + 2)(1))

(x + 2)4

( X + 2)2 – ( X – 5)(2x + 4)

(x + 2)4

x2 + 4x + 4 – (2×2 – 6x – 20)

(x + 2)4

x2 + 4x + 4 – 2×2 + 6x + 20

(x + 2)4

-x2 + lOx + 24

(x + 2)4

– (x2 – lOx – 24)

(x + 2)4

-(x + 2)(x – 12)

(x + 2)4

-(x – 12)

(x + 2)3

Note: Alternatively, we could have factored the term x + 2 from both the numerator and denominator immediately after using the Quotient Rule like we have done previously to simplify the derivative. There are a variety of techniques you can use to algebraically manipulate a function correctly!

Now, we find the x-values where .f'(x) = 0 or J'(x) does not exist. J'(x) = 0 when the numerator equals 0, and .f'(x) does not exist when the denominator equals zero (remember we are looking at the domain of .f’ when determining where it does not exist). First, we will find the x-values where .f'(x) = 0:

J'(x) = 0

-(x – 12) = 0

x=12

Next, to find the x-values where .f'(x) does not exist, we set the denominator equal to

zero:

J'(x) DNE

(x + 2)3 = 0

((x + 2)3)31

= (0)3

x+2=0

X= -2

Because f (x) =(:;i)2 is a rational function and its denominator must be nonzero, its domain is (-oo, -2) U (-2, oo). Thus, only x = 12 is in the domain, so it is the only critical value of f.

Graphing the function initially to observe its behavior on the interval (-4, 15), we may see a graph that looks similar to the one shown in Figure 3.3.22:

Figure 3.4.22: Graph of the function f(x) =(:;i)2

Looking at this graph, it appears as though there is no absolute minimum on the interval (-4, 15) because of the vertical asymptote at x = -2. It also looks like there may not be an absolute maximum due to the function appearing to get closer and closer to the horizontal asymptote y = 0.

Because we know there may be some type of extrema at x = 12 because it is a critical value of .f, we need to adjust our calculator’s viewing window so we can see the behavior of the function near this critical value. See Figure 3.4.23.

Figure 3.3.23: Graph of the function f(x) = (:;2)2 near x = 12

We can see from this graph that the function crosses the x-axis and there is an absolute maximum. Because we know x = 12 is the only critical value of f, the absolute maximum must occur at x = 12. Recall that we could see the vertical asymptote at x = -2 when we first graphed f (see Figure 3.4.22 again), so we know there is no absolute minimum.

To find the absolute maximum off on the interval (-4, 15) exactly, we calculate f(12):

x-5

f(x) = (x+ 2)2

. 12 – 5 7

j (12) = (12+ 2)2 = 196 0.0357

Thus, the absolute maximum of f on the interval (-4, 15) 1s absolute minimum.

7 but there 1s no

Note: f does have a horizontal asymptote, !I = 0, but it reaches its absolute maximum at

x = 12 before approaching the line!/= 0.

Try It 7

Find the absolute maximum and minimum of the function f(x) = 8 / x^2 – 4 on each of the following intervals, if they exist.

(a) [0, 5]

(b) (-1, 1)

(c) [-6, -3)

Other Tests for Absolute Extrema

Recall that we learned two tests for finding local extrema: the First Derivative Test (Section 3.1) and the Second Derivative Test (Section 3.2). In certain circumstances, these tests can also be used to find absolute extrema.

If a function, f, has only one critical value in some interval on which f is continuous and there is a local extremum at that critical value, then the local extremum must also be an absolute extremum. Why should this make sense intuitively? If a continuous function has a local maximum at x = 2, for example, and it has no other critical values other than x = 2, then the graph of the function would never be able to “turn around” and increase so that it reaches a value greater (higher) than that at x = 2. Thus, the local maximum would also be the absolute maximum.

Figure 3.4.24 shows the graph of a continuous function with only one critical value, x = 2, that has a local maximum at the critical value. Because x = 2 is the only critical value of f, the graph of the function will never “turn around” again. So the local maximum is also the absolute maximum:

Figure 3.3.24: Graph of a function f that has both a local and absolute maximum at x=2

We will use this result in the next section when solving optimization problems!

Try It Answers

1. Local maximum of 16 at x = 2; Local mm1mum of -16 at x

maximum; No absolute minimum

-2; No absolute

2. No local maxima; Local minimum of -8 at x

minimum is -8 at x = 2

2; No absolute maximum; Absolute

3. (a) Absolute maximum: 11; Absolute minimum: -16

(b) Absolute maximum: 16; Absolute minimum: -16

4. (a) Absolute maximum at x = -5; No absolute minimum

(b) No absolute maximum; No absolute minimum

(c) Absolute maximum at x = 5; Absolute minimum at x = 2

5. (a) Absolute maximum: 1 9; Absolute minimum: 57

(b) Absolute maximum: 168; Absolute minimum: 57

6. (a) Absolute maximum: ½; Absolute minimum: 1

(b) Absolute maximum: 2l 6; Absolute minimum: 3\/’3

5 2

7. Absolute maximum: None; Absolute minimum: None

8. Absolute maximum: -2; Absolute minimum: None

9. Absolute maximum: None; Absolute minimum: ¼

Exercises

Basic Skills Practice

For Exercises 1 – 5, the graph off is shown. Find (a) the x-values of any local extrema and

(b) the x-values of any absolute extrema off. Specify whether an extremum is a minimum or maximum.

1.

2.

3.

-“j’- y

-6- –

-5- —

-4- –

-3- —

-2- –

-1- –

4.

5.

For Exercises 6 – 11, the graph off is shown. Find the absolute maximum and minimum of

f on each of the given intervals, if they exist.

6.

I I I I I I I I I I I I I I I

0 0 0 . .–. I I I I I I I I I I I

L—‘—…0—“—-L L.

(a) [2, 6] (b) [-3, 4.5] (c) [5,7]

7.

(a) [-1,2] (b) [-9, 8] (c) [-7,5]

8.

(a) [-4, 2] (b) [-5,5] (c) [-4, 4.5]

9.

-.4

‘

(a) [-3, 2] (b) [-9, 7] (c) [2,8]

10.

– — — :- 2 — ‘ — ,— ,— ,— -, —,—-, —‘ — ‘ — -‘ — ‘ — ‘ — – ,—‘ — -,— ,—-, , —

J I I I I I I I I I J I I 1 I I I I I

– – – – – – -f 1:5 – –f- – – – – – – – – – – – -:- – –:- – – – – – – — -}- – – t- – -7- – – – – -.J,- — – – – -:- – – -:- – –:- – – – – –

I I I I I I I I

– — —+–1· —+—-:—-:—-:—-:—-:—-:—-r—r—+—+—1—1—1—-:—-:—-:—-:—

, I I I I I I I I I I I I I I I I I I

– — — 0:5 — — — — —: —‘—- — —t— —t— — — —:– ———x–

• 1 , , 1 , 2 : 3 , 4 : 5 : fl , 7 : 8 , 9

– – – – – – -6:5 —f- – – – – – _JI_- – — – -:- – –:– – – – – – — – – – -T- – – – – – – – – – — – – – -:- – – -:- – – -:- – – – – –

I I I I I I I I I I I I I I I I I I I I

– —: — : –t — :— :— :— :—-:—-:—-: —: —: —: —: — :— :— :—-:—-:—-:—-: —

‘ I I I I I I I I I I I I I I I I I I I

I I I I I I I I

(a) (2,6) (b) [3,8) (c) (1, 6]

11.

(a) (-1.4, 1.4) (b) [O, 1) (c) (-2,-1]

For Exercises 12 – 16, use the Closed Interval Method to find the absolute maximum and minimum of f on each of the given intervals.

12. f(x) = x5 + 5×4 – 35×3

(a) [-9, 7] (b) [-1,4] (c) [-8, 6]

Intermediate Skills Practice

For Exercises 22 – 24, the graph of f is shown. Find (a) any local extrema and (b) any absolute extrema, as well as where they occur, of f. Specify whether an extremum is a minimum or maximum.

22.

23.

24.

I, I, ,I ,I ,I I, I, I, —-rI —–,I ,I 1I I,

11

:

‘ : ‘ :

‘

‘ ‘

‘ ‘ ‘ : ‘

:

x:

For Exercises 25 and 26, the graph of f is shown. Find the absolute extrema of f on each of the given intervals, if they exist.

25.

:

(a) (-8,-4) (b) [O, 6) (c) [1,3]

26.

(a) [-3,2] (b) (-1, 5) (c) [2,8)

For Exercises 27 – 34, find the absolute maximum and minimum of f on each of the given intervals.

27. J(x) = (x – 2)ex

(a) [O, 2]

(b)

[-5, O]

(c)

[1,3]

28.J(x) = (x2 – 4)5

(a) [-4,-1](b)[-1,3](c)[-5, 5]

29. f(x) = ln(:)

X

(a) [1,3](b)[1/2, ye](c)[2,4]

30.

J(x) = i(x2 – 1)2

(a) [-2, 3](b)[-1,4](c)[-1, 1]

31.J(x) = e-x2

(a) [-7, -3](b)[-1,1](c)[0,3/2]

32.J(x) = 4×2 ln(x)

(a) [1, 3](b)[2,5](c)[1/10, 3/4]

x3

33. f(x) = –

x2 -9

(a) [-15, -10]

34. J(x)=x-5ln(x)

(a) [1, 6]

(b) [-2,1] (c) [4,6]

(b) [2, 4] (c) [15,20]

For Exercises 35 – 40, find the absolute maximum and minimum off on each of the given intervals, if they exist.

4x

35. f x( )= x2 – 4x+ 4

(a) [1, 6]

36. J(x) = {/(x2 – 3x)4

(a) [-4,1)

37. f (X ) = – 3x

X +4

(a) (-8, 4)

-5ex

38. f(x) = -2x –

(a) (-1,3)

39. J(x) = ln(x) – 2×2

(a) [1/3, 8/9)

x2 1

40. f(x)

X – 25

(a) [-4,3)

(b) [-8, 0) (c) (-2,4]

(b) (2,5) (c) (0,3]

(b) (-2, 2] (c) [0,6)

(b) (1, 4] (c) (5,8]

(b) (1/2, 4) (c) (0,2]

(b) [-7,2] (C) (1, 7]

Mastery Practice

41. Given the graph off shown below, find (a) any local extrema and (b) any absolute extrema, as well as where they occur, off. Specify whether an extremum is a minimum or maximum.

42. Given the graph off shown below, find the absolute extrema off on each of the given intervals, if they exist.

(a) (-6,0]

(b) [-3,6]

(c) [0,3)

(d) (-oo, oo)

(e) (3,6]

(f) (-3, 0)

For Exercises 43 – 52, find the absolute maximum and minimum of f on each of the given intervals, if they exist.

(a) (10, 60)

5

(b) [-1,52] (c) (-25,-15)

44. f(x) =(

x )

x+6

(a) [-4, 8]

45. J(x) = 2x ln(x)

(a) [1/4, 3/4)

(b) [-2,3) (c) (-8,-5]

(b) (0, 1] (c) (1, 5)

e0.25×2

46. J(x) = —

x

(a) [-4, -2] (b) [1/2, 2] (c) [-2, 2]

47. f(x) = 3×4 – 40×3 + 150×2 – 25

(a) (-2, 5)

(x + 5)4

(b) [1,6] (c) (-4, -1]

48. f(x) =(

x-l )3

(a) [-6,20]

49. j(x) = (x – 1)10(x – 5)10

(a) [1/2, 2]

16- x2

50. J(x) = .

X +9

(a) [-3,3]

f( ) 4ln(x2)

X

(a) (1, 5)

52. f(x) = i(x3 – 3x)4

(a) [-3,2]

(b) [2,15) (c) [-7,0]

(b) (3/2, 11/2) (c) [1,3]

(b) (-1, 2) (c) (0,4]

(b) [-4, 2] (c) [1, e)

(b) (2, 5) (c) [-3/2, 0)

53. Given the graph of g shown below, find (a) where any local extrema and (b) where any absolute extrema off occur. Specify whether an extremum is a minimum or maximum.

54. Determine where each of the following functions has an absolute minimum, if it exists, on the interval (0, oo).

-5×24

(a) J(x) = —

e3x

2×14

(b) g(x) = e

For Exercises 55 and 56, use the given information to sketch a possible graph of f.

55.

• The domain of f is (-oo, oo).

• f is discontinuous at x = 7 only.

• f has a local and absolute minimum

at X = 7.

• f does not have local or absolute maxima.

56.

• The domain of f is [0, 4].

• f is discontinuous at x = 4 only.

• f has an absolute minimum at x = 0.

• f has an absolute maximum at x = 4.

• f does not have any local extrema.

Communication Practice

57. Is it possible for a function to have a local maximum at x

absolute extrema? Explain.

6, but not have any

58. Is it possible for f to have a local minimum at x = 3 if f is discontinuous at x = 3? Explain.

59. Explain, in general terms, the Extreme Value Theorem.

60. Where on the graph of a function can absolute extrema occur?

61. Briefly describe the three steps of the Closed Interval Method.

62. To use the Closed Interval Method, why does the function need to be continuous and the interval closed?

63. Can a function have more than one absolute maximum? Explain.

64. Are we able to use the Closed Interval Method to find the absolute minimum of f(x) =

;,r;._5 on the interval [-7, -1]? Explain.

65. If we are unable to use the Closed Interval Method to find the absolute extrema of a function on an interval because the function is not continuous or the interval is not closed (or both), how should we proceed in order to find any absolute extrema?

66. Compare and contrast local and absolute extrema.

3.4 Optimization

In this section, we will apply the calculus techniques we have learned thus far to find the optimal solution of real-world problems. For instance, a manufacturing company may want to maximize its profit or minimize its cost. A farmer may need to build a pen for his cattle with maximum area, but he may have a limited amount of fencing with which to build the pen.

An optimal solution is the absolute maximum (if we are maximizing a quantity) or the ab solute minimum (if we are minimizing a quantity) of a function. The function we are trying to maximize or minimize is called the objective function. Because we will be working with mostly real-world problems, there will be constraints (due to limited resources, etc.) asso ciated with the objective function. Thus, we will have to consider these constraints when trying to find the optimal solution.

Rather than graphing the objective function on our calculator and approximating the abso lute maximum or minimum, we will use calculus techniques developed throughout Chapter 3 so we can find the exact, optimal solution!

Learning Objectives:

In this section, you will learn how to solve real-world optimization problems using calculus techniques. Upon completion you will be able to:

Create variables and formulate equations to develop an objective function that must be maximized or minimized and represents a “numbers” optimization problem.
Create variables and formulate equations to develop an objective function that must be maximized or minimized and represents a business optimization problem.
Create variables and formulate equations to develop an objective function that must be maximized or minimized and represents a geometric optimization problem.
Create realistic intervals corresponding to the variable used during the optimization process.
Apply the appropriate calculus technique (Closed Interval Method, First Derivative Test, or Second Derivative Test) while performing the optimization process to calculate an optimal solution.

Optimization Process

The process of finding an absolute maximum or minimum is called optimization. Again, the function we are maximizing or minimizing (i.e., optimizing) is called the objective function. The quantity we are optimizing, which is given by the objective function, can be recognized by its proximity to “est” words (greatest, smallest, largest, least, most, farthest, highest, etc.), or the problem may explicitly say to maximize or minimize a certain quantity. For

example, we may need to find the dimensions of a cardboard box with the largest possible volume given some constraints pertaining to the amount of cardboard available. In this case, our objective function would be a function representing the volume of the box because this is the quantity we would need to maximize.

Often, the most challenging part of solving an optimization problem is understanding the situation at hand and translating it into mathematical form. To translate the problem into mathematical form, we have to create variables and understand the relationships between the variables to formulate the objective function (it can be helpful to draw a picture, when applicable). Also, if there are constraints given in the problem, we have to translate those into mathematical form as well. We will create constraint equations that will relate the variables in the objective function. The constraint equations are always equations, so they should have equal signs.

Usually, an objective function will consist of two (or more) variables. If an objective function has two variables and there is a constraint equation associated with the problem, we can solve the constraint equation for one of the variables and then rewrite the objective function as a function of just one variable. This will be the goal in most of our problems!

Because we will be solving real-world optimization problems, we will also have to find an interval on which the objective function makes sense in terms of the context of the prob lem (i.e., find the domain of the objective function). For example, a company cannot sell a negative number of items, but it is possible that it will not sell any items. The same with the price per item: The price cannot be negative, but it is possible the company may be giving items away for free in order to advertise their product. Another example would be as mentioned above in the introduction: If a farmer is building a pen to hold cattle, the dimensions of the pen must be positive. Creating an interval on which the objective function is defined ensures we are considering the realistic aspects of the problem.

After setting up the optimization problem by finding the objective function and correspond ing interval, we will use the Closed Interval Method we learned in Section 3.4 to find the optimal solution if we are maximizing or minimizing an objective function on a closed interval and the function is continuous on the interval. If, instead, we are maximizing or minimizing an objective function on an open interval in which the function is continuous and it has only one critical value in the interval, we can use either the First Derivative Test (Section 3.1) or the Second Derivative Test (Section 3.2) to determine the optimal solution (assuming the necessary conditions for each test are met, which in this textbook, they will be). You may be wondering how we can use these tests for local extrema to find absolute extrema. We mentioned this briefly at the end of Section 3.4: If a function has only one critical value in some interval on which the function is continuous and there is a local extremum at that crit ical value, then the local extremum must also be an absolute extremum because the graph of the function will not be able to “turn around” and reach a point that is higher (or lower) than the local extremum.

After solving an optimization problem using calculus, we need to go back and reread the problem to be sure that we answer the question that is asked. It is possible that we may

forget the original question in the problem after performing the entire optimization process! We now summarize the steps for solving an optimization problem:

Optimization Process for Solving Optimization Problems

1. Translate the word problem into mathematical form:

• Identify the quantity you want to maximize or mm1m1ze as well as the quantity you know (typically given by a constraint).

• Introduce variables and draw a picture with labels, if applicable.

• Determine the relationships between the variables to create the objective function and, possibly, constraint equation. Remember a constraint equa tion should have an equal sign and be equal to a number!

2. State the objective function in terms of one variable. If there is more than one variable in the objective function, solve the constraint equation for one variable and substitute the result into the objective function. This will give an objective function with only one variable.

3. Find the interval on which the objective function must be optimized. To do this, consider the context of the problem and restrict the variables to determine a realistic domain for the objective function. 4. Use calculus to find the optimal solution:

• Find the critical values of the objective function. Recall these are the x values where f'(x) = 0 or f'(x) does not exist, and they are in the domain (i.e., in the interval found above in step 3).

• If the interval in step 3 is closed and the objective function is continuous on the interval, use the Closed Interval Method to find the absolute maximum or minimum. Remember, this means to evaluate the objective function at the critical values in the interval as well as the endpoints of the interval.

• If the interval in step 3 is open and there is only one critical value in the interval, perform one of the following tests to find the absolute maximum or mm1mum:

the First Derivative Test (assuming the objective function is continuous on the interval and is differentiable near, and on both sides of, the critical value)

the Second Derivative Test (assuming the objective function is twice differentiable at the critical value and the first derivative equals zero at the critical value)

Remember, if a function has only one critical value in an interval, the func tion is continuous on the interval, and a local extremum occurs at the critical value, then the local extremum must also be an absolute extremum.

4. Reread the problem to be sure you answer the original question.

Note: If the objective function is twice differentiable on the interval, then all the necessary conditions to use either the First Derivative Test or the Second Derivative test are satisfied. Recall from Section 2.2 that if a function is differentiable on an interval, then it must be continuous on the interval. And, if a function is twice differentiable on an interval, then its first derivative certainly exists on the interval. In this textbook, the objective functions we formulate will be twice differentiable on their respective intervals. Hence, we can use either the First Derivative Test or the Second Derivative Test to find the absolute extremum when the objective function has only one critical value in an open interval.

In this textbook, we will focus on solving three types of optimization problems: “numbers”, business, and geometric problems.

“Numbers” Problems

We will start with a “numbers” optimization problem to help demonstrate the optimization process. A “numbers” problem usually consists of a question that asks us to find two numbers that meet certain criteria. We will see this in our first example.

Examplel Find two nonnegative numbers x and y such that 3x + y = 30 and their product is a maximum.

Solution: The word “maximum” tells us this is an optimization problem, so we proceed with the optimization process outlined previously:

1. Translate the word problem into mathematical form:

To translate the problem, we express the quantities mathematically we want (the objective function) and know (in this case, a constraint equation). Because we want the product of the numbers to be a maximum, the objective function will represent the product of the numbers. We will call the function P to represent the product:

want: P = xy

The equation given in the problem is what we know, so it is the constraint equa tion. Recall constraint equations always have equal signs:

know: 3x + y = 30

2. State the objective function in terms of one variable:

Because the objective function has two variables, x and y, we must solve the constraint equation for one variable and substitute the result into the objective function. We can choose either x or y to solve for in the constraint equation, but we will choose y because it will be slightly easier than solving for x:

3x + y = 30

y = 30 – 3x

Substituting y into the objective function gives

P=xy

= x(30 – 3x)

= 30x – 3×2

Thus, the objective function in terms of one variable, x, is

P(x) = 30x – 3×2

3. Find the interval on which the objective function must be optimized:

To find the interval, we must consider the context of the problem and restrict the variables to determine a realistic domain for the objective function. However, in this particular problem, we are explicitly given restrictions: both x and y are nonnegative (meaning they must equal zero or be positive). So we have

x 2: 0 and y 2: 0

However, because the objective function is in terms of x, we need to substitute for y in the second inequality to ensure our restrictions lead us to an interval consisting of x-values. Recall that y = 30 – 3x from step 2. Substituting for 1/ in the second inequality and solving for x gives

y2′.0==}

30 – 3x 2: 0

30 2: 3x

10 2: x, or x ::; 10

Thus, we must maximize the objective function, P(x) = 30x-3×2, on the interval [0, 10]. Note that the interval has brackets because the inequalities were inclusive, and therefore the interval is closed.

4. Use calculus to find the optimal solution:

To find the optimal solution, we need to find the critical values of the objective function, P, we found in step 2 and use an appropriate test that we learned previously to find the absolute maximum.

The critical values of Pare thex-values where P'(x) = 0 or P'(x) does not exist, and they are in the domain of P (i.e., they are in the interval we found in step 3). To find the critical values of P, we must find P'(x) first:

P(x) = 30x – 3×2 ==}

P'(x) = 30 – 6x

Now, we find the x-values where P'(x) = 0 or P'(x) does not exist. Because P’ is a polynomial and has a domain of all real numbers, it will exist everywhere. Thus, we only need to find the x-values where P'(x) = 0:

P'(x) = 0 ==}

30 – 6x = 0

30 = 6x

5=x

Notice that x = 5 is in the interval [0, 10] we found in step 3. Therefore, x = 2 is a critical value of P.

Because the interval [0, 10] is closed and the objective function, P, is continuous on the interval, we will use the Closed Interval Method we learned in section 3.4 to find the absolute maximum. Because we already know the critical value, x = 5, we can go straight to evaluating the objective function at the critical value and the endpoints of the interval:

P(x) = 30x – 3×2

P(5) = 30(5) – 3(5)2 = 75

P(O) = 30(0) – 3(0)2 = 0

P(lO) = 30(10) – 3(10)2 = 0

Looking at the function values, we see the largest is 75 at x

absolute maximum of P is 75, and it occurs at x = 5.

5. Reread the problem to be sure you answer the original question:

5. Thus, the

After working through the optimization process, it is easy to forget what the original question asked us to find! Looking back at the problem, we see that we need to find two numbers, x and y, whose product is a maximum. We found that the maximum product is 75, and it occurs at x = 5. So we have one number: x = 5. Thus, we just need to find the other number, y.

To find y, we substitute x = 5 into the constraint equation we solved for yin step

2:

y = 30-3x

y = 30 – 3(5)

= 15

Therefore, the numbers whose product is a maximum are x = 5 and y = 15.

Try It 1

Find two nonnegative numbers x and y whose sum is 100 and their product is a maximum.

Business Problems

We will now turn our attention to business optimization problems. These types of optimiza tion problems usually consist of a question in which we are trying to maximize revenue or profit.

Example 2 A concert promoter has found that if she sells tickets for $50 each, she can sell 1200 tickets. For each $5 she raises the price, 50 less people attend. For what price should she sell each ticket to maximize her revenue? What is her maximum revenue?

Solution: We need to maximize revenue, so we know this is an optimization problem. We will proceed with the optimization process:

1. Translate the word problem into mathematical form:

To translate the problem, we express the quantities mathematically we want (the objective function) and know. Because we want the revenue to be a maximum, the objective function will be the revenue function.

Although we are not given a constraint equation we know in this example, there are several pieces of information we know that will help us create the objective function in terms of one variable (we are given information about ticket price and demand). Because the objective function will represent revenue, we will call the function R and apply the general formula for revenue:

want: R = x • p

where x is the number of tickets sold at a price of $p per ticket.

Again, we do not have a constraint equation given, but we know that when price is $50 per ticket, 1200 tickets will be sold. Also, we know that for every $5 increase in price, 50 fewer tickets are sold. We can use this information to find the price-demand function, p.

VVe know p is a linear function because the problem says for every $5 increase in price, 50 fewer tickets are sold. We can find the equation of the line by finding the slope of the line and then using the point-slope formula to find the equation.

To find the slope of the line, we need two points. We can use the information in the problem to get two points of the form (x,p). We know one point is (1200, 50), and we can create another point, (1150, 55), based on the information about price and demand. Now, we can calculate the slope, m:

m, =

P2 -pi

X2 – X1

==}

m,=

55- 50

1150 – 1200

5

-50

1

10

Note: Recall that you can let either point be (x1, p1) or (x2, p2) when calculating the slope, m. You will get the same slope no matter which points you consider to be (x1, p1) and (x2, p2). Now, we will use the point-slope formula to find the equation of the line:

P – P1 = m (x – x1) ==}

p – 50 =

1

–(x – 1200)

10

1

p- 50 = –x + 120

10

1

p = –x+ 170

10

Note: Again, you can use either point in the point-slope formula above, and you will get the same answer.

Thus, we can now say we know the price-demand function, p:

know: p(x) = –

1 x + 170

10

2. State the objective function in terms of one variable:

Substituting the function we know, p, into the objective function we want, R =

x • p, we get

R(x) = x • p(x)

= X ( – /oX + l 70)

= –x1 2 + 170x 10

Thus, the objective function in terms of one variable, x, is

3. Find the interval on which the objective function must be optimized:

To find the interval when we are working business optimization problems, we have to remember that the objective function is based on the variables x and p (quantity and price, respectively). Both of these quantities must be nonnegative (meaning they must equal zero or be positive). Remember, there may not be any items sold (concert tickets in this case), or the items may be given away for free for advertising purposes. So in terms of restrictions, we have

x 2:: 0 and p(x) 2:: 0

Recall that p(x) = – /0x + 170 from step 1. Substituting for p(x) in the second inequality and solving for x gives

p(x) 2:: 0

–x1 + 170 > 0

10 –

170 ->

1

-1×0

1700 2:: x, or x ::::; 1700

Thus, we must maximize the objective function, R(x) = -/0x2 + l 70x, on the

interval [0, 1700]. Note that the interval is dosed, as it should be for business

optimization problems!

4. Use calculus to find the optimal solution:

To find the optimal solution, we need to find the critical values of the objective function, R, we found in step 2 and use an appropriate test that we learned previously to find the absolute maximum.

The critical values of Rare the x-values where R'(x) = 0 or R'(x) does not exist, and they are in the domain of R(x) (i.e., they are in the interval we found in step 3). To find the critical value of R, we must find R'(x) first:

R(x) = –

1 x2 + 170x ==;,

10

R'(x) = – 2

1

x + 170

0

Now, we find the x-values where R'(x) = 0 or R'(x) does not exist. Because R’ is a polynomial and has a domain of all real numbers, it will exist everywhere. Thus, we only need to find the x-values where R'(x) = 0:

R'(x) = 0 ==;,

2

–x + 170 = 0

10

2

170 = lOX

1700 = 2x

850 = X

Notice that x = 850 is in the interval [O, 1700] we found in step 3. Therefore,

x = 850 is a critical value of R.

Because the interval [O, 1700] is closed and the objective function, R, is continuous on the interval, we will use the Closed Interval Method we learned in Section 3.4 to find the absolute maximum. Because we already know the critical value, x = 850,

we can go straight to evaluating the revenue function at the critical value and the endpoints of the interval:

1 .

R(x) = – x2 + l 70x ==;,

10

R(850) = –

1 (850)2 + 170(850) = $72, 250

1 0

R(O) = -/ (0)2 + 170(0) = $0

0

R(l 700) = -/ (1700)2 + 170(1700) = $0

0

Looking at the function values, we see the largest is $72,250 at x = 850. Thus, the maximum revenue is $72, 250, and it occurs when 850 concert tickets are sold.

5. Reread the problem to be sure you answer the original question:

Looking back at the problem, we see that we need to find the selling price of each ticket that will maximize the concert promoter’s revenue. We found that the maximum revenue is $72,250, and it occurs at x = 850 (i.e., when 850 tickets are sold). However, we need to find the selling price of each ticket, which is given

by the price-demand function p we found in step 1. We will calculate p(850) to determine the selling price:

p(x) =

1

–x

10

1

+ 170 ==}

p(850) = -10 (850) + 170

= $85

Therefore, the concert promoter must charge $85 per ticket to maximize her rev enue. The maximum revenue is $72, 250.

Example 3 A company sells x ribbon winders per year at a price of $p per ribbon winder. The company’s price-demand function for the ribbon winders is given by p(x) = 300-0.02x.

The ribbon winders cost $30 each to manufacture, and the company has fixed costs of $9000 per year. How many ribbon winders must the company sell to maximize its profit? What is the company’s maximum profit?

Solution: vVe need to maximize profit, so we proceed with the optimization process:

1. Translate the word problem into mathematical form:

To translate the problem, we express the quantities mathematically we want (the objective function) and know. Because we want the profit to be a maximum, our objective function will be the profit function.

Although we are not given a constraint equation we know in this example, there are several pieces of information we know that will help us create the objective function in terms of one variable (we are given the price-demand function and information about cost). Because the objective function will represent profit, we will call the function P and apply the general formula for profit:

want: P=R-C

where R and C represent the revenue and cost functions, respectively.

Again, we do not have a constraint equation given, but we do know the price demand function, p, and we can use it to find the revenue function, R (remember R = x • p, where xis the number of items sold at a price of $p per item). We can also use the information about cost to find the cost function, C. Then, we can find the objective function, P.

First, let’s find the revenue function, R, by substituting the price-demand func tion, p, given to us in the problem:

R(x) = x • p(x)

= x(300 – 0.02x)

= 300x – 0.02×2

Thus, we now know the revenue function R(x) = 300x – 0.02×2, where R(x) 1s the revenue, in dollars, from selling x ribbon winders:

know: R(x) = 300x – 0.02×2

Next, we must find the cost function, C. The general formula for cost is given by C = V x + F, where x is the number of items made, V represents the cost of producing each item, and F represents the company’s fixed costs. The problem

states that it costs 30 to make each ribbon winder and that the company has fixed costs of $9000 per year. Thus, we know the cost function is given by C(x) = 30x + 9000, where C(x) is the cost, in dollars, of making x ribbon winders:

know: C(x) = 30x + 9000

2. State the objective function in terms of one variable:

Substituting the functions we know, Rand C, into the objective function we want,

P = R- C, gives

P(x) = R(x) – C(x)

= (300x – 0.02×2) – (30x + 9000)

= (300x – 0.02×2) – 30x – 9000)

= -0.02×2 + 270x – 9000

Thus, the objective function in terms of one variable, x, is

P(x) = -0.02×2 + 270x – 9000

3. Find the interval on which the objective function must be optimized:

Like the previous business example, the objective function is based on the variables

x and p (quantity and price, respectively), and they must both be nonnegative.

So in terms of restrictions, we have

x 2 0 and p(x) 2 0

Recall that p(x) = 300 – 0.02x. Substituting for p(x) in the second inequality and solving for x gives

p(x) 2 0 ==}

300 – 0.02x 2 0

300 2 0.02x

15, 000 2 x, or x ::; 15, 000

Thus, we must maximize the objective function, P(x) = -0.02×2 + 270x – 9000, on the interval [O, 15,000]. Again, note that the interval is closed.

4. Use calculus to find the optimal solution:

To find the optimal solution, we need to find the critical values of the objective function, P, we found in step 2 and use an appropriate test that we learned previously to find the absolute maximum.

The critical values of Pare thex-values where P'(x) = 0 or P'(x) does not exist, and they are in the domain of P (i.e., they are in the interval we found in step 3). To find the critical values of P, we must find P'(x) first:

P(x) = -0.02×2 + 270x – 9000

P'(x) = -0.04x + 270

Now, we find the x-values where P'(x) = 0 or P'(x) does not exist. Because P’ is a polynomial and has a domain of all real numbers, it will exist everywhere. Thus, we only need to find the x-values where P'(x) = 0:

P'(x) = 0

-0.04x + 270 = 0

270 = 0.04x

6750 = X

Notice that x = 6750 is in the interval [O, 15,000] we found in step 3. Therefore,

x = 6750 is a critical value of P.

Because the interval [O,15,000] is closed and the objective function, P, is continu ous on the interval, we will use the Closed Interval Method we learned in Section

3.4 to find the absolute maximum. Because we already know the critical value, x = 6750, we can go straight to evaluating the profit function at the critical value and the endpoints of the interval:

P(x) = -0.02×2 + 270x – 9000

P(6750) = -0.02(6750)2 + 270(6750) – 9000 = $902,250

P(O) = -0.02(0)2 + 270(0) – 9000 = -$9000

P(15, 000) = -0.02(15, 000)2 + 270(15, 000) – 9000 = -$459, 000

Looking at the function values, we see the largest is $902, 250 at x = 6750. Thus, the maximum profit is $902,250, and it occurs when 6750 ribbon winders are sold.

5. Reread the problem to be sure you answer the original question:

Looking back at the problem, we see that we need to find how many ribbon winders the company must sell to maximize its profit. In other words, we need the quantity, which is x. We found that the absolute maximum occurs at x = 6750.

Thus, the company must sell 6750 ribbon winders to maximize its profit. We also found the maximum profit in step 4: $902,250.

Notice in each of the examples above, the optimal solution occurred at the critical value and not at an endpoint of the interval. It is possible that an optimal solution may occur at an endpoint, so we must always check the function values at the endpoints as well as the critical value(s).

Try It 2

A company that makes toy firetrucks as a price-demand function given by p(x) = -0.02x + 45, where x is the number of trucks sold at a price of $p each . The company has a weekly cost function, in dollars, given by C(x) = 12x+ 1000, where x is the number of trucks manufactured each week. Determine the selling price of each truck that will maximize the company’s profit.

Geometric Problems

Finally, we will learn how to use the optimization process to find optimal solutions to problems involving geometry. We will work several examples to demonstrate this type of opti mization problem because they tend to be the most challenging. And, unlike the business optimization problems, the intervals associated with geometric optimization problems are typically open. We will see why in the next example!

Example 4 A rectangular garden is to be constructed using an existing rock wall for one side and wire fencing for the other three sides. If 100 feet of wire fencing is to be used, determine the dimensions of the garden so it has the largest area possible. What is the largest area?

Solution: We need to find the largest area possible, which means we need to maximize area. So we proceed with the optimization process:

1. Translate the word problem into mathematical form:

To translate the problem, we express the quantities mathematically we want (the objective function) and know (in this case, a constraint equation). Because we want the area to be a maximum, the objective function will give the area of the garden.

Although we are not explicitly given a constraint equation we know, we do know there is only 100 feet of wire fencing available. We will be able to use this known value to create a constraint equation.

Often when we have a geometric optimization problem, it is helpful to draw a picture, define variables, and label the picture in order to formulate the objective function and constraint equation. The garden is rectangular, so we will draw a rectangle and designate one of the walls to be the rock wall (it does not matter which one!). Next, we will define y to be the length, in feet, of the side opposite the rock wall. Finally, we will let x be the length, in feet, of each of the remaining two sides. See Figure 3.4.1.

Figure 3.4.1: Rectangular garden with a rock wall on one side

The objective function must represent the area of the garden because we want to maximize its area. The area of a rectangle is equal to its width times its length. We will create the objective function using our variables, and we will call the objective function A because it gives the area of the garden:

want: A= xy

Vve can also create the constraint equation using our picture. We know 100 feet of wire fencing will be used, so we can create an equation we know representing the amount of fencing used. We will add the lengths of all the sides, except for the rock wall, and set the sum equal to the total amount of wire fencing (recall constraint equations have equal signs):

know: x + y + x = 100

2x + y = 100

2. State the objective function in terms of one variable:

Because the objective function has two variables, x and y, we must solve the constraint equation for one variable and substitute the result into the objective function. We can choose either x or y to solve for in the constraint equation, but we will choose y because it will be slightly easier than solving for x:

2x + y = 100

y = 100 – 2x

Substituting y into the objective function gives

A=xy

= x(lO0 – 2x)

= l00x – 2×2

Thus, the objective function in terms of one variable, x, is

A(x) = lO0x – 2×2

3. Find the interval on which the objective function must be optimized:

To find the interval, we must consider the context of the problem and determine a realistic domain for the objective function. When solving geometric optimization problems, we need to remember that dimensions must be positive. Why? They certainly cannot be negative, and if they equal zero there would be no geometric shape (in this case, there would be no garden!). In this example, the dimensions are given by x and y, so we have

x > 0 and ‘!J > 0

However, because the objective function is in terms of x, we need to substitute for y in the second inequality to ensure our restrictions lead us to an interval consisting of x-values. Recall that y = 100 – 2x from step 2. Substituting for y in the second inequality and solving for x gives

y > O=;, 100 – 2x > 0

100 > 2x

50 > X, or X < 50

Thus, we must maximize the objective function, A(x) = l00x-2×2, on the interval (0, 50). Note that the interval has parentheses because the inequalities were strict inequalities. Therefore, the interval is open, which it typically is for geometric optimization problems!

4. Use calculus to find the optimal solution:

To find the optimal solution, we need to find the critical values of the objective function, A, we found in step 2 and use an appropriate test to find the absolute maximum.

The critical values of A are the x-values where A’ ( x) = 0 or A’ (x) does not exist, and they are in the domain of A (i.e., they are in the interval we found in step 3). To find the critical values of A, we must find A'(x) first:

A(x) = lO0x – 2×2 =}

A'(x) = 100 – 4x

Now, we find the x-values where A'(x) = 0 or A'(x) does not exist. Because A’ is a polynomial and has a domain of all real numbers, it will exist everywhere. Thus, we only need to find the x-values where A'(x) = 0:

A'(x) = 0 =;, 100 – 4x = 0

100 = 4x

25 = X

Notice that x = 25 is in the interval (0, 50) we found in step 3. Therefore, x = 15 is a critical value of A (and the only critical value in the interval).

Because the interval (0, 50) is open, we cannot use the Closed Interval Method to find the absolute maximum. However, because the objective function only has one critical value in the interval and is twice differentiable on the interval, we can use either the First Derivative Test we learned in Section 3.1 or the Second Derivative Test we learned in Section 3.2 (because the critical value occurs when A'(x) = 0) to find the absolute extremum.

Recall from Section 3.4 that if a function has only one critical value in some interval on which the function is continuous and there is a local extremum at that critical value, then the local extremum must also be an absolute extremum because the graph of the function will not be able to “turn around” and reach a point that is higher (or lower) than the local extremum. As stated previously, either the First Derivative Test or the Second Derivative Test for local extrema can be used to find absolute extrema when working geometric optimization problems in this textbook because the objective functions will only have one critical value in and be twice differentiable on their corresponding (open) intervals.

In this example, we will demonstrate using the First Derivative Test. Because we already know the critical value, x = 25, we can go straight to creating the sign chart of A'(x) to determine the behavior of the objective function at the critical value.

Recall from Section 3.1, we place the critical value (also a partition number) on a number line and indicate that it is in the interval with a solid dot. We will also place a dotted line at x = 0 and x = 50 on the number line so we remember not to select x-values to test that are outside of the interval (0, 50). Now, we need to determine the sign of A'(x) on the intervals (0, 25) and (25, 50). We will choose the x-values x = 10 and x = 30 to test:

A'(x) = 100 – 4x ==}

A'(lO) = 100 – 4(10) = 60 > 0

A'(30) = 100 – 4(30) = -20 < 0

Using this information, we can fill in the sign chart of A’ (x). Because we are also interested in the information this yields for A, we include that information below the number line. See Figure 3.4.2.

Figure 3.4.2: Sign chart of A'(x) with the corresponding information for

A(x) = lO0x – 2×2

Using Theorem 3.1, we see that a local maximum occurs at x = 25 (note that we also double check that A is defined at x = 25, which it is). Because x = 25 is the only critical value in the interval (0, 50) and A is continuous on the interval, we

know this local maximum is also the absolute maximum. Now, we need to double check that we are in fact looking for an absolute maximum, which we are!

5. Reread the problem to be sure you answer the original question:

Looking back at the problem, we see that we need to find the dimensions that will maximize the area of the garden as well as the maximum area. The dimensions are given by x and y. We found that the absolute maximum occurs at x = 25

feet, and we can find y by substituting x = 25 into the constraint equation we

solved for y in step 2:

y = 100- 2x ==}

y = 100 – 2(25)

= 50 feet

To find the maximum area, we substitute x = 25 into the objective function:

A(x) = lO0x – 2×2 ==}

A(25) = 100(25) – 2(25)2 = 1250 square feet

Therefore, the dimensions that maximize the area of the garden are 25 feet by 50 feet. The maximum area is 1250 square feet. Note that the length of the rock wall is 50 feet.

Example 5 A box with a square base and an open top must have a volume of 32,000 cubic cen timeters. Find the dimensions of the box so that the least amount of material is used to make the box.

Solution: This problem asks us to find the least amount of material used to make the box, which means we need to minimize the surface area of the box. So we proceed with the optimization process:

1. Translate the word problem into mathematical form:

To translate the problem, we express the quantities mathematically we want (the objective function) and know (in this case, a constraint equation). Because we want the surface area to be a minimum, the objective function will give the surface area of the box.

Although we are not explicitly given a constraint equation we know, we do know that the box must have a volume of 32,000 cubic centimeters. We will be able to use this known value to create a constraint equation. Because we have a geometric optimization problem, we will draw a picture, define variables, and label the picture to help us create the objective function and constraint equation. We will define x to be the length and width, in centimeters, of the bottom of the box. Remember, the box has a square base. We will let y be the height, in centimeters, of the box. See Figure 3.4.3.

Figure 3.4.3: Box with a square base and open top

The objective function must represent the surface area of the box (i.e., the amount of material used to make the box) because we want to minimize the surface area. The surface area of the box is equal to the sum of the areas of all its surfaces. We can create the objective function using our variables, and we will call the objective function S because it gives the surface area. Adding the areas of all the surfaces, except for the top of the box because it is open, we can formulate the objective function we want. We start by finding the area of the bottom of the box and then adding the areas of all four sides:

S = xx +xy+xy+xy +x

want: S = x2 + 4xy

We can also create the constraint equation using our picture. We know the volume is 32,000 cubic centimeters, so we can create an equation we know representing the volume of the box. The volume of the box is equal to its length times its width times its height:

know: x • x • y = 32, 000

x2y = 32,000

2. State the objective function in terms of one variable:

Because the objective function has two variables, x and y, we must solve the constraint equation for one variable and substitute the result into the objective function. vVe can choose either x or y to solve for in the constraint equation, but we will choose y because it will be easier than solving for x:

x2y = 32,000

32,000

·y=

x2

Substituting y into the objective function gives

S = x2 + 4xy

= X 2 + 4X

(32,000)

x2

128,000

=X +

X

Thus, the objective function in terms of one variable, x, is

S(x)= x2+ 128,000

X

3. Find the interval on which the objective function must be optimized:

To find the interval, we must consider the context of the problem and determine a realistic domain for the objective function. We know the dimensions of the box must be positive. In this example, the dimensions are given by x and y, so we have

x > 0 and y > 0

However, because the objective function is in terms of x, we need to substitute for y in the second inequality to ensure our restrictions lead us to an interval consisting of x-values. Recall that y = 32 00 from step 2. Substituting for y in the second inequality gives

y > O==;, 32,000

2 > 0

X

Unlike our previous examples, we cannot solve this inequality algebraically for x (if we multiply both sides by x2, we lose the variable!). Thus, we have to simply observe this quotient and note that the numerator is positive and the denominator will always be positive as long as x =J. 0 because x is squared. Hence, this quotient will always be positive provided x =J. 0.

Thus, we have x > 0 and x =J. 0, which means the only restriction is x > 0. Therefore, we must maximize the objective function, S(x) = x2 + 128;000, on the interval (0, oo). Note that the interval is open.

How can we have an interval in which the x-value can be infinitely large when there is a restricted volume of 32,000 cubic centimeters? Remember, this interval applies to x-values only. So if x is a really large number, the box can still have a volume of 32,000 cubic centimeters as long as y is a really small number!

4. Use calculus to find the optimal solution:

To find the optimal solution, we need to find the critical values of the objective function, S, we found in step 2 and use an appropriate test to find the absolute minimum.

The critical values of Sare the x-values where S'(x) = 0 or S'(x) does not exist, and they are in the domain of S (i.e., they are in the interval we found in step 3). To find the critical values of S, we must find S'(x) first. Before doing so, we will rewrite the function S to avoid using the Quotient Rule:

S( X ) = X 2 + 128, 000

X

= x2 + 128, ooox-1

Finding the derivative and simplifying gives

S'(x) = 2x – 128, ooox-2

= 2x-

128,000

x2

To find the critical values of S, we find the x-values where S'(x) = 0 or S'(x) does not exist. First, let’s find the x-values where S'(x) = 0:

S'(x) = O

2x _ 128,000= 0

x2

2x= 128,000

x2

(x2) (2x)= 128, 00 (x2)

X

2×3 = 128, 000

x3 = 64 000

‘

X = 40

Now, we find the x-values where S'(x) does not exist. S'(x) = 2x- 12! 00 does not exist when x2 = 0 (remember we are looking at the domain of S’ when determining where it does not exist). Solving the equation x2 = 0 for x gives x = 0. Hence, S'(x) does not exist when x = 0.

Note: Recall that, technically, S’ (x) also does not exist when x < 0 because we restricted the domain (i.e., interval) of S to be (0, oo). However, as discussed previously, we would not consider these to be “important” x-values when finding where S'(x) does not exist. Furthermore, we are looking for the critical values of S, and these x-values would not be critical values because they are not in the restricted domain (i.e., interval) of S. Therefore, the only “important” x-value where S'(x) does not exist is x = 0.

Of the x-values x = 40 and x = 0, only x = 40 is in the interval (0, oo). Thus, it is the only critical value of S.

Because the interval (0, oo) is open, we cannot use the Closed Interval Method to find the absolute minimum. However, because the objective function only has one critical value in the interval and is twice differentiable on the interval, we can use either the First Derivative Test or the Second Derivative Test (because the critical value occurs when S'(x) = 0) to find the absolute extremum. Because we reviewed the First Derivative Test in the last example, we will use the Second Derivative Test in this example.

We already know the critical value, x = 40, so we can go straight to finding S11 x) and evaluating it at the critical value. To avoid using the Quotient Rule, we will use the original form of S’ (x) to find S11( x):

S'(x) = 2x – 128, ooox-2

S”(x) = 2 + 256, ooox-3

Because we only have to find the value of S”(x) when x = 40 to determine if there is an absolute minimum, there is no need to continue simplifying the second derivative. We now find S” (40):

S”(x) = 2 + 256, ooox-3 ==}

S”(40) = 2 + 256, 000(40)-3 = 6 > 0

Using Theorem 3.4, we see that S is concave up at x = 40, so there is a local minimum at x = 40. Because x = 40 is the only critical value in the interval (0, oo) and Sis continuous on the interval, this local minimum is also the absolute minimum of the objective function on the interval (which is good because our goal was to minimize the surface area!).

5. Reread the problem to be sure you answer the original question:

Looking back at the problem, we see that we need to find the dimensions that will minimize the surface area of the box. In other words, we need both x and y. We found that the absolute minimum occurs at x = 40 centimeters, and we can find y by substituting x = 40 into the constraint equation we solved for y in step

2: 32,000

y = ==}

x2

32,000

y = (40)2

= 20 centimeters

Therefore, the dimensions that minimize the surface area of the box (and, hence, the amount of material used to make the box) are 40 centimeters by 40 centime ters by 20 centimeters. Note that the length and width of the box are each 40 centimeters, and the height of the box is 20 centimeters.

Try It 3

Kirby has 1728 square inches of material, and he needs to make a box with a square base and an open top that has the largest volume possible. Find the dimensions and volume of this box.

In some cases, we may have a geometric optimization problem in which we are asked to minimize cost (so in a sense, this is a business application). However, the solution will follow the ideas we have established for solving a geometric optimization problem, as we will see in the next example.

Example 6 Farmer Ben needs to enclose a rectangular pen that has a total area of 2560 square feet and a partition dividing the pen in order to separate his chickens and pigs (see Figure 3.4.4). The fencing to build the partition, which needs to be reinforced to keep the animals in their respective regions, costs $24 per foot. The rest of the fencing needed to build the pen costs $8 per foot. Find the length of the partition that will minimize Farmer Ben’s cost of building the pen.

Figure 3.4.4: Rectangular pen with a partition

Solution: This problem asks us to find the length of the partition that will minimize Farmer Ben’s cost. We proceed with the optimization process:

1. Translate the word problem into mathematical form:

To translate the problem, we express the quantities mathematically we want (the objective function) and know (in this case, a constraint equation). Because we want the cost to be a minimum, the objective function will give the total cost of building the pen with the partition.

Although we are not explicitly given a constraint equation we know, we do know the total area of the pen is 2560 square feet. We will be able to use this known value to create a constraint equation.

Because we have a geometric optimization problem, we will draw a picture, define variables, and label the picture to help us create the objective function and the constraint equation. We will define x to be the length, in feet, of the partition and the sides parallel to the partition. We will let y be the length, in feet, of the remaining sides. See Figure 3.4.5.

Figure 3.4.5: Rectangular pen with a partition and variables x and ·y labeled

The objective function will represent the cost of all the fencing needed to enclose the pen with the partition because we want to minimize cost. We must express the cost of each side of the pen and the partition using our variables and the cost per foot, and then we add the expressions to get the total cost. \Ne will call the objective function C because it represents cost. Adding the cost of each side and then the partition gives

want:

C = 8x + 8x + 8y + 8y + 24x

C = 40x + l6y

We can also create the constraint equation using our picture. We know the to tal area of the pen is 2560 square feet, so we can create an equation we know representing the area. The area of the pen is equal to its width times its length:

know: xy = 2560

2. State the objective function in terms of one variable:

Because the objective function has two variables, x and y, we must solve the constraint equation for one variable and substitute the result into the objective function. We can choose either x or y to solve for in the constraint equation, but we will solve for y:

xy = 2560

2560

y=–

x

Substituting ‘!} into the objective function we get

C = 40x + l6y

= 40x + 16 (25×60)

= 40x + 40′.960

X

Thus, the objective function in terms of one variable, x, is

C(x)= 40x+ 40,960

X

3. Find the interval on which the objective function must be optimized:

To find the interval, we must consider the context of the problem and determine a realistic domain for the objective function. Again, the dimensions of the pen must be positive. In this example, the dimensions are given by x and ‘!}, so we have

x > 0 and’!}> 0

However, because the objective function is in terms of x, we need to substitute for y in the second inequality to ensure our restrictions lead us to an interval consisting of x-values. Recall y = 2 60 from step 2. Substituting for y in the second inequality gives

y>O=;,

2560> 0

X

Like the previous example, we cannot solve this inequality algebraically for x (if we multiply both sides by x, we lose the variable!). Thus, we have to simply observe this quotient and note that the numerator is positive and the denominator will be positive as long as x > 0. Therefore, the quotient itself will be positive when x > 0.

So the only we restriction we have is that x > 0.

Thus, we must maximize the objective function, C(x)

interval (0, oo). Note that the interval is open.

4. Use calculus to find the optimal solution:

40x + 40’960 on the

To find the optimal solution, we need to find the critical values of the objective function, C, we found in step 2 and use an appropriate test to find the absolute maximum.

The critical values of Care the x-values where C'(x) = 0 or C'(x) does not exist, and they are in the domain of C (i.e., they are in the interval we found in step 3). To find the critical values of C, we must find C'(x) first. Before doing so, we will rewrite the function C in order to avoid using the Quotient Rule:

C(x=)

40x+ 40,960

X

= 40x + 40, 960x-1

Finding the derivative and simplifying gives

C'(x) = 40 – 40, 960x-2

= 40-

40,960

x2

To find the critical values of C, we find the x-values where C'(x) = 0 or C'(x)

does not exist. First, let’s find the x-values where C'(x) = 0:

C'(x) = O

40 _ 40, 960= O

x2

40= 40,960

x2

(x2) (40) = 40 60 (x2)

40×2 = 40, 960

x2 = 1024

x = -32 and x = 32

Because the dimensions of the pen must be positive, we can disregard the negative x-value. Hence, C'(x) = 0 when x = 32.

Now, we find the x-values where C'(x) does not exist. C'(x) = 40 – 40 60 does not exist when x2 = 0 (remember we are looking at the domain of C’ when determining where it does not exist). Solving the equation x2 = 0 for x gives x = 0. Hence, C'(x) does not exist when x = 0.

Note: Recall that, technically, C'(x) also does not exist when x < 0 because we restricted the domain (i.e., interval) of C to be (0, oo). However, as discussed previously, we would not consider these to be “important” x-values when finding where C'(x)

does not exist. Furthermore, we are looking for the critical values of C, and these x-values would not be critical values because they are not in the restricted domain (i.e., interval) of C. Therefore, the only “important” x-value where C'(x) does not exist is x = 0.

Of the x-values x = 32 and x = 0, only x = 32 is in the interval (0, oo). Thus, it is the only critical value of C.

Because the interval (0, oo) is open, we cannot use the Closed Interval Method to find the absolute minimum. However, because the objective function only has one critical value in the interval and is twice differentiable on the interval, we can again use either the First Derivative Test or the Second Derivative Test (because

the critical value occurs when C'(x) = 0) to find the absolute extremum. We will

use the First Derivative Test in this example.

Because we already know the critical value, x = 32, we can go straight to creating the sign chart of C’ (x) to determine the behavior of the objective function at the critical value.

We will place the critical value (also a partition number) on a number line, and indicate that it is in the interval with a solid dot. We will also place a dotted line at x = 0 on the number line so we remember not to select x-values to test that are outside of the interval (0, oo).

Now, we need to determine the sign of C'(x) on the intervals (0, 32) and (32, oo). We will choose the x-values x = 20 and x = 40 to test:

C'(x)= 40 _ 40,960

x2

C'(20) = 40 – 40,960 = -62.4 0

(20)2 <

C'(40) = 40 – 40,960 = 14.4 O

(40)2 >

Using this information, we can fill in the sign chart of C’ (x). Because we are also interested in the information this yields for C, we include that information below the number line. See Figure 3.4.6.

Figure 3.4.6: Sign chart of C'(x) with the corresponding information for C(x) = 40x + 40′;60

Using Theorem 3.2, we see that a local minimum occurs at x = 32 (note that we also double check that C is defined at x = 32, which it is). Because x = 32 is the only critical value in the interval (0, oo) and C is continuous on the interval, we

know this local minimum is also the absolute minimum. Now, we double check that we are in fact looking for an absolute minimum, which we are!

5. Reread the problem to be sure you answer the original question:

Looking back at the problem, we see that we need to find the length of the partition that will minimize Farmer Ben’s cost. In other words, we only need the value of x. We found that the absolute minimum occurs at x = 32 feet, so the partition needs to be 32 feet long to minimize cost.

Example 7 Zachary needs to construct a box whose length is twice its width and has a volume of 50 cubic feet. The material used to build the top and bottom of the box costs $10 per square foot, and the material used to build the sides costs $6 per square foot. Find Zachary’s minimum cost of constructing the box.

Solution: This problem asks us to minimize the cost of constructing the box, so we proceed with the optimization process:

1. Translate the word problem into mathematical form:

To translate the problem, we express the quantities mathematically we want (the objective function) and know (in this case, a constraint equation). Because we want the cost to be a minimum, the objective function will give the cost of con structing the box.

Although we are not explicitly given a constraint equation we know, we do know the volume of the the box is 50 cubic feet. We will be able to use this known value to create a constraint equation.

Because we have a geometric optimization problem, we will draw a picture, de fine variables, and label the picture to help us create the objective function and constraint equation. We will define x to be the width, in feet, of the box. This means the length, in feet, of the box will be 2x because the length is twice the width. We will let h be the height, in feet, of the box. See Figure 3.4.7.

Figure 3.4.7: Box whose length is twice its width

The objective function must represent the cost of all the materials needed to construct the box because we want to minimize cost. We must express the cost of each side of the box using our variables and then add the expressions to get

the total cost. Note that the cost of each side is given by its area, in square feet, multiplied by its respective cost per square foot.

We will call the objective function C because it represents cost. Adding the cost of materials to make each side of the box (starting with the top and bottom which cost $10 per square foot), we get

C = lO(x • 2x) + lO(x • 2x) + 6(x • h) + 6(2x • h) + 6(x • h) + 6(2x • h) ===}

want: C = 20×2 + 20×2 + 6xh + l2xh + 6xh + l2xh ===}

C = 40×2 + 36xh

We can also create the constraint equation using our picture. We know the volume of the box is 50 cubic feet, so we can create an equation we know representing the volume. The volume of the box is equal to its length times its width times its height:

know:

2x • x • h = 50 ===}

2x2h = 50

2. State the objective function in terms of one variable:

Because the objective function has two variables, x and h, we must solve the constraint equation for one variable and substitute the result into the objective function. We can choose either x or h to solve for in the constraint equation, but we will solve for h because it is easier:

2×2 h = 50 ===}

h = 50= 25

2×2 x2

Substituting h into the objective function gives

C = 40×2 + 36xh

= 40×2 + 36x (! )

= 40×2+ 900

X

Thus, the objective function in terms of one variable, x, is

2 900

C( x) =40x+

x

3. Find the interval on which the objective function must be optimized:

To find the interval, we must consider the context of the problem and determine a realistic domain for the objective function. The dimensions of the box must be positive, or there will be no box! In this example, the dimensions are given by x, 2x, and h, so we have

x > 0, 2x > 0, and h > 0

Notice that solving the second inequality, 2x > 0, for x gives x > 0, which we already have from the first inequality. However, because the objective function is in terms of x, we need to substitute for h in the third inequality to ensure our restrictions lead us to an interval consisting of x-values. Recall h = ; from step

2. Substituting for h in the second inequality gives

h>0

25

– >0

x2

Like the previous examples, we cannot solve this inequality algebraically for x (if we multiply both sides by x2, we lose the variable!). Thus, we have to simply observe this quotient and note that the numerator is positive and the denominator will always be positive as long as x =/=- 0 because x is squared. Hence, this quotient will always be positive provided x -/:- 0.

Thus, we have x > 0 and x =/=- 0, which means the only restriction is x > 0. Therefore, we must maximize the objective function, C(x) = 40×2 + 9 0, on the interval (0, oo). Note that the interval is open.

4. Use calculus to find the optimal solution:

To find the optimal solution, we need to find the critical values of the objective function, C, we found in step 2 and use an appropriate test to find the absolute maximum.

The critical values of Care the x-values where C'(x) = 0 or C'(x) does not exist, and they are in the domain of C (i.e., they are in the interval we found in step 3). To find the critical values of C, we must find C’ (x) first. Before doing so, we will rewrite the function C in order to avoid using the Quotient Rule:

C(x) = 40×2 + -900

X

= 40×2 + 1800x-1

Finding the derivative and simplifying gives

C'(x) = 80x – 900x-2

= 80x – 900

x2

To find the critical values of C, we find the x-values where C'(x) = 0 or C'(x)

does not exist. First, let’s find the x-values where C'(x) = 0:

C'(x) = 0 ==;,

80x – 900 = 0

x2

80x= 900

x2

(x2) 80x = 9:20 (x2) 80×3 = 900

3 45

X =-

==;, X = if¥ 2.2407

Now, we find the x-values where C'(x) does not exist. C'(x) = 80x – 9i2°does not exist when x2 = 0 (remember we are looking at the domain of C’ when determining where it does not exist). Solving the equation x2 = 0 for x gives x = 0. Hence, C'(x) does not exist when x = 0.

Note: Recall that, technically, C'(x) also does not exist when x < 0 because we restricted the domain (i.e., interval) of C to be (0, oo). However, as discussed previously, we would not consider these to be “important” x-values when finding where C'(x) does not exist. Furthermore, we are looking for the critical values of C, and these x-values would not be critical values because they are not in the restricted domain (i.e., interval) of C. Therefore, the only “important” x-value where C’ (x) does not exist is x = 0.

Of the x-values x = efij- and x = 0, only x = efij- is in the interval (0, oo). Thus, it is the only critical value of C.

Because the interval (0, oo) is open, we cannot use the Closed Interval Method to find the absolute minimum. However, because the objective function only has one critical value in the interval and is twice differentiable on the interval, we can again use either the First Derivative Test or the Second Derivative Test (because the critical value occurs when C'(x) = 0) to find the absolute extremum. We will use the Second Derivative Test in this example.

Because we already know the critical value, x = efij-, we can go straight to finding C” (x) and evaluating it at the critical value. To avoid using the Quotient Rule, we will use the original form of C’ (x) to find C” (x):

C'(x) = 80x – 900x-2

C”(x) = 80 + 1800x-3

Because we only have to find the value of C”(x) when x = efij- to determine if there is an absolute minimum, there is no need to continue simplifying the second

derivative. We now find C” ( ifij-):

C” ( fi) 80 + 1800 ( fi)-‘ 210 > 0

Using Theorem 3.4, we see that C is concave up at x = ifij-,so there is a local minimum at x = ifij-. Because x = ifij- is the only critical value in the interval

(0, oo) and C is continuous on the interval, this local minimum is also the absolute minimum of the objective function on the interval (which is good because our goal was to minimize the cost!).

5. Reread the problem to be sure you answer the original question:

Looking back at the problem, we see that we need to find the minimum cost of constructing the box. We know the absolute minimum occurs at x = ifij- feet,

so we need to evaluate the cost (objective) function we found in step 2 at this x-value:

2

C ( 3(45) = 40 ( 3(45)

+ 900 $602.49

V4 V4 if¥

Thus, Zachary’s minimum cost of constructing the box is $602.49.

Try It 4

A manufacturer is going to make rectangular storage containers with open tops. The volume of each container must be 10 cubic meters, and the length of each container will be twice its width. The cost of the material for the four sides of the container is $4 per square meter. The cost of the material for the bottom of the container is $8 per square meter. Find the manufacturer’s minimum cost of making each container.

Try It Answers

l. X = 50; y = 50

2. $28.50

3. 24 inches by 24 inches by 12 inches; 6912 cubic inches

4. $115.86

Exercises

Basic Skills Practice

1. Find two nonnegative numbers whose sum is 40 and their product is a maximum.

2. Find two numbers whose difference is 16 and their product is a minimum.

3. Find two positive numbers whose product is 30 and their sum is a minimum.

4. The revenue function for Star Bright Sneakers is given by R(x) = 208x -0.2×2 dollars, where x is the number of pairs of sneakers sold. How many pairs of sneakers must the company sell to maximize its revenue?

5. Reid’s Pizzeria has a daily profit function given by P(x) = -60 + 12x – ½x2 dollars, where x is the number of pizzas sold. How many pizzas must her pizzeria sell to maximize profit?

6. Cut That Grass! sells a popular lawn mower and has determined its revenue function is given by R(x) = -.04×2 +350x dollars, where xis the number of lawn mowers sold.

(a) How many lawn mowers must the company sell to maximize revenue?

(b) What is the maximum revenue?

7. Shake It Up, an appliance company that sells high-end blenders, has a monthly profit function given by P(x) = -0.3×2 + 450x – 10,500 dollars, where x is the number of blenders sold.

(a) How many blenders must the company sell to maximize profit?

(b) What is the maximum profit?

8. You have 400ft of fencing to construct a rectangular pen for cattle. What are the dimensions of the pen that maximize its area?

9. You need to construct a fence around a rectangular area of 1600 square feet. What are the dimensions of the pen that will minimize the amount of material needed to construct the fence?

10. Roy Lee, a famed ostrich farmer, wants to enclose a rectangular area and then divide it into five pens with fencing parallel to one side of the rectangle as shown below. He will use 420 feet of fencing to complete the job.

(a) Find the dimensions of each pen that will maximize the total area of all five pens.

(b) \i\That is the largest possible total area of all five pens?

11. The manager of a garden store wants to build a 600 square foot rectangular enclosure on the store’s parking lot in order to display some equipment. Three sides of the enclosure will be built of redwood fencing, at a cost of $7 per foot. The fourth side will be built of cement blocks, at a cost of $14 per foot. Find the dimensions of the enclosure that will minimize cost.

12. You are building five identical pens adjacent to each other such that they have a total area of 1000 m2 as shown in the figure below. Find the dimensions of each pen that will minimize the amount of fencing needed. Round your answers to three decimal places, if necessary.

13. Bob needs to fence in a right-angled triangular region that will border a (mostly straight) river as shown below. The fencing for the left border costs $9 per foot, and the lower border costs $3 per foot. Bob doesn’t need any fencing along the side of the river. He has $630 to spend, and he wants as much area as possible.

(a) Find the dimensions of the region that will provide the largest possible area.

Hint: The area of a triangle is ½• base • height.

(b) What is the largest possible area of the region?

Intermediate Skills Practice

14. Find two nonnegative numbers x and y whose product is 250 and 2x+5y is minimized.

15. Find two positive numbers x and y such that x + y = 10 and x2y2 is maximized.

16. Find two numbers x and y such that x + y = 15 and x2 – y is minimized.

17. Super Rad Bikez specializes in selling glow in the dark bikes for kids. It has determined its price-demand function to be p(x) = 200- 0.8x, where xis the number of bikes sold at a price of $p each. Find the number of bikes the company must sell to maximize its

revenue.

18. A curling iron manufacturer has revenue and cost functions, both in dollars, given by R(x) = 35x – 0.lx2 and C(x) = 4x + 2000, respectively, where x is the number of curling irons made and sold.

(a) How many curling irons must the company sell to maximize its profit?

(b) What is the company’s maximum profit?

19. The price-demand and cost functions, both in dollars, for a company that makes and sells cordless drills are given by p(x) = 143 – 0.04x and C(x) = 75,000 + 75x, respec tively, where x is the number of cordless drills made and sold.

(a) Find the price per drill that will maximize profit.

(b) Find the cost when profit is maximized.

20. When Mrs. Stewart, a theater owner, charges $5 dollars per ticket, she sells 250 tickets. Through market analysis, the owner learns that for every one dollar she raises the price, she will lose 10 customers. What price should she charge for each ticket to maximize revenue? 21. The total cost, in dollars, for Marsha to make x oven mitts is given by

C(x) = 64 + l.5x + O.Olx2.

(a) Find a function that gives the average cost per oven mitt, C(x).

(b) How many oven mitts should Marsha make to minimize her average cost?

21. Mrs. Barker has 400 feet of fencing to enclose a rectangular vegetable garden. What should the dimensions of her garden be in order to enclose the largest area?

22. Ainzley has 600 feet of fencing available to construct a rectangular pen with a fence divider down the middle of the pen. What are the dimensions of the pen that will enclose the largest total area?

23. Brian has 800ft of fencing to build a pen for his hogs. If he has a river on his property that he can use for one side of the pen, what is the maximum area of the pen?

24. Mrs. Savage wants to enclose a rectangular garden. The area of the garden is 800 square feet. The fence along three sides costs $6 per foot, and the fence along the fourth side, which needs to be reinforced to keep out her cows, costs $18 per foot. Find the length of the reinforced fence that will minimize Mrs. Savage’s cost of enclosing the garden.

25. Angus has $240 dollars to buy fencing material to enclose a rectangular plot of land. The fencing for the north and south sides costs $6 per foot, and the fencing for the east and west sides costs $3 per foot.

(a) Find the dimensions of the plot with the largest area.

(b) For this largest plot, how much money will be used to build the east and west sides?

26. A box with a square base and an open top must have a volume of 216in3. Find the dimensions of the box so that the least amount of material is used to make the box. Round your answers to three decimal places, if necessary.

27. Coker is constructing a box for his cat to sleep in. The plush material for the square bottom of the box costs $5/ft2, and the material for the sides costs $2/ft2. Coker needs a box with a volume of 4ft3. Find the dimensions of the box that will minimize cost. Round your answers to three decimal places, if necessary.

28. Kirby has 1875 cm2 of material, and he needs to make a box with a square base and an open top that has the largest volume possible.

(a) Find the dimensions of his box.

(b) “\i\That is the volume of his box?

29. Fred is going to make rectangular storage containers. Each container will have a square base and a volume of 128 cubic feet. The cost of material for the top and four sides is

$2 per square foot, and the cost of material for the bottom is $6 per square foot. Find the minimum cost for which Fred can make each storage container.

30. Sarah Jo needs to construct a box without a top whose length is twice its width and whose volume is 72ft3. The material used to build the bottom of the box costs $8/ft2, and the material used to build the sides costs $4/ft2. Find the dimensions of the box that will minimize Sarah Jo’s cost of constructing the box.

31. You need to form a 10-inch by 15 -inch piece of tin into a box (without a top) by cutting a square from each corner of the tin and folding up the sides. Round your answers below to three decimal places, if necessary.

(a) Find the dimensions of the box with the largest volume.

(b) What is the largest possible volume?

32. U.S. postal regulations state that the sum of the length and girth (distance around) of a package must be no more than 108 inches. Find the dimensions of an acceptable box that has the largest volume possible if its end is a rectangle whose length is twice its width.

Mastery Practice

34. You must construct a box with a square base and an open top that will hold 100 cubic inches of water. Round your answers below to three decimal places, if necessary.

(a) Find the dimensions of the box that will require using the least amount of material.

(b) “\i\That is the least amount of material needed to make the box?

35. Suppose the box in the previous exercise (Exercise #35) is made of different materials for the bottom and the sides. If the bottom material costs $0.05 per square inch and the side material costs $0.03 per square inch, what is the cost of the least expensive box that will hold 100 cubic inches of water?

36. Find two nonnegative numbers whose sum is 36 and the sum of their squares 1s a minimum.

37. Suppose you decide to create a rectangular garden in the corner of your yard and enclose it with fencing. Two sides of the garden will be bounded by your existing fence, so you only need to enclose the other two sides. You have 80 feet of fencing. Find the maximum area of your new garden.

38. A rectangular box with a square bottom and closed top is to be made from two mate rials. The material for the sides costs $1.50 per square foot, and the material for the top and bottom costs $3.00 per square foot. If you are willing to spend $15 on the box, what is the largest volume it can contain? Round your answer to three decimal places, if necessary.

39. A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y = 3 – x2. What are the dimensions of such a rectangle with the greatest possible area?

40. Party Tune, a company that makes sing-a-long microphones for kids, has found that it can sell 500 microphones each month when the price of a microphone is $86. For every one dollar decrease in price, Party Tune can sell 5 additional microphones each month.

(a) Find the price Party Tune must charge per microphone to maximize its revenue.

(b) “\i\That is the maximum revenue?

41. A rectangular storage container with an open top is to have a volume of 28 cubic meters. The length of its base is twice the width. Material for the base costs 10 dollars per square meter. Material for the sides costs 7 dollars per square meter. Find the cost of materials for the cheapest such container.

42. An open box is to be made from a 10-inch by 16-inch piece of cardboard by cutting out squares of equal size from the four corners and folding up the sides.

(a) Find the length and width of each square so that the volume of the box is maxi mized.

(b) “\i\That is the largest volume of the box?

43. Find two positive numbers x and y such that x + ·y = 10 and ·y – .!. is a maximum. 44. The price-demand function for a certain flashlight is given by p(x) = 15-0.00lx, where x is the number of flashlights sold at a price of $p each. The flashlight manufacturer has fixed costs of $10, 000 per month, and it costs $3.00 to make each flashlight.

(a) How many flashlights must be sold to maximize the manufacturer’s profit?

(b) What is the maximum profit?

44. The top and bottom margins of a poster are 2 cm long, and the side margins are each 4 cm long. If the area of printed material on the poster is fixed at 384 cm2, find the dimensions of the poster with the smallest area. Round your answers to three decimal places, if necessary.

45. A farmer wants to start ra1smg cows, horses, goats, and sheep, and desires to have a rectangular pasture for the animals to graze in. However, no two different kinds of animals can graze together. In order to minimize the amount of fencing she will need, she has decided to enclose a large rectangular area and then divide it into four equally-sized pens by adding three segments of fencing inside the large rectangle that are parallel to two existing sides. She has decided to purchase 7500ft of fencing to use. What is the maximum possible area that each of the four pens will enclose?

46. Vicki makes and sells backpack danglies. The cost, in dollars, for Vicki to make x danglies is given by C(x) = 75 + 2x + 0.015×2. Find Vicki’s minimum average cost for making danglies.

47. Heather needs to construct a box whose length is three times its width and whose volume is 50 cubic feet. The material used to build the top and bottom of the box costs $10 per square foot, and the material used to build the sides costs $6 per square foot. What is the minimum cost of the box?

48. You have 100 feet of fencing to build a pen in the shape of a circular sector (see the “pie slice” shown below). The area of such a sector is given by A = r;, where r is the radius of the circle and s is the arc length of the sector. What value of r maximizes the enclosed area?

49. Find the positive integer that minimizes the sum of the number and its reciprocal.

50. A small business that sells wind chimes can sell 100 wind chimes each month when the price per wind chime is $25. For every dollar the price is raised, 10 fewer wind chimes will be sold. If the business has a cost function given by C (x) = 9x + 1000 dollars,

where x is the number of wind chimes produced, find the revenue of the business when its profit is maximized.

51. Find the length of the cut, x, to make a box with the largest volume formed from a 10-inch by 10-inch piece of cardboard cut and folded as shown below.

52. You own a small airplane that holds a maximum of 20 passengers. It costs you $100 per flight from St. Thomas to St. Croix for gas and wages plus an additional $6 per passenger for the extra gas required by the extra weight. The charge per passenger is $30 each if 10 people charter your plane (10 is the minimum number you will fly), and this charge is reduced by $1 per passenger for each passenger over 10 who travels. What number of passengers on a flight will maximize your profit?

53. You have been asked to determine the least expensive route for a telephone cable that connects Allenville with Boontown. The towns are separated by a river that is 2 km wide, and the horizontal (land) distance between them is 8 km (see the figure below). If it costs $5000 per mile to lay the cable on land and $8000 per mile to lay the cable across the river (with the cost of the cable included), find the cost of the least expensive route. Round your answer to the nearest dollar.

54. Dr. Whitfield claims the “productivity levels” of people in various fields can be de scribed as a function of their “career age” t by p(t) = e-at – e-bt, where a and b are constants depending on the field, and career age is approximately 20 less than the

actual age of the individual. Based on this model, at what career ages do (a) mathe maticians (a= 0.03, b = 0.05), (b) geologists (a= 0.02, b = 0.04), and (c) historians (a= 0.02, b = 0.03) reach their maximum productivity? Round your answers to the nearest year.

55. You have T dollars to buy fencing material to enclose a rectangular plot of land. The fencing for the north and south sides costs $A per foot, and the fencing for the east and west sides costs $B per foot. Find the dimensions of the plot with the largest area.

Communication Practice

57. Briefly describe the five steps of the optimization process.

58. When using the optimization process to solve an optimization problem involving geo metric regions, you should translate the problem by first doing what?

59. Under what circumstance is a constraint equation required to solve an optimization problem?

60. After finding the interval on which to maximize or minimize an objective function in step 2 of the optimization process, describe the calculus techniques that can be used to solve the problem in step 3 as well as the condition(s) for using each.

61. If a continuous function has only one critical value, x = 4, and a local maximum at

x = 4, explain why the function also has an absolute maximum at x = 4.

13 Chapter 13

Chapter 3

Curve Sketching and Optimization

3.3 Absolute Extrema

Identifying Absolute Extrema Graphically

Absolute Extrema on an Interval

Determining Absolute Extrema Using Calculus

Other Tests for Absolute Extrema

Exercises

Basic Skills Practice

Intermediate Skills Practice

Mastery Practice

Communication Practice

3.4 Optimization

Optimization Process

“Numbers” Problems

Business Problems

Geometric Problems

Try It Answers

Exercises

Basic Skills Practice

Intermediate Skills Practice

Mastery Practice

Communication Practice

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