CC Applied Optimization

Section 3.3 Applied Optimization

Motivating Questions

In a setting where a situation is described for which optimal parameters are sought, how do we develop a function that models the situation and then use calculus to find the desired maximum or minimum?

Supplemental Videos.

The main topics of this section are also presented in the following videos:

How to Solve It
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unl.yuja.com/V/Video?v=7114310&node=34303272&a=117261420&autoplay=1
Initial Example
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unl.yuja.com/V/Video?v=7114312&node=34303290&a=91796962&autoplay=1
Advanced Example
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unl.yuja.com/V/Video?v=7114313&node=34303302&a=116662963&autoplay=1

Near the conclusion of Section 3.2, we considered two optimization problems in which determining the function to be optimized was part of the problem. Example 3.23 sought to maximize the total area enclosed by the combination of an equilateral triangle and a square built from a single piece of wire (cut in two). Example 3.26 subsequently investigated how the volume of a box constructed by removing squares from the corners of a piece of cardboard is dependent on the size of the squares removed.

Neither of these problems explicitly provided a function to optimize. Instead, we first tried to understand the problem by drawing a figure and introducing variables, and then sought to develop a formula for a function that modeled the quantity to be optimized. Upon establishing a function, we chose an appropriate domain and then were finally ready to apply the ideas of calculus to determine the absolute minimum or maximum.

In this section the primary emphasis is on the reader solving problems. Initially, some substantial guidance is provided, but the problems progress to require greater independence as we move along.

Example 3.27.

According to U.S. postal regulations, the girth plus the length of a parcel sent by mail may not exceed 108 inches, where by “girth” we mean the perimeter of the smallest end. What is the largest possible volume of a rectangular parcel with a square end that can be sent by mail? What are the dimensions of such a package? The process of answering these questions has been broken up into a sequence of tasks:

Let $x$ represent the length of one side of the square end and $y$ the length of the longer side. Label these quantities appropriately on an image like the one shown in Figure 3.28 below.

Figure 3.28. A rectangular parcel with a square end.
What is the quantity to be optimized in this problem? Find a formula for this quantity in terms of $x$ and $y\text{.}$
The problem statement tells us that the parcel’s girth plus length may not exceed 108 inches. In order to maximize volume, we assume that we will actually need the girth plus length to equal 108 inches. What equation does this produce involving $x$ and $y\text{?}$
Solve the equation you found in (c) for one of $x$ or $y$ (whichever is easier).
Now use your work in (b) and (d) to determine a formula for the volume of the parcel so that this formula is a function of a single variable.
Over what domain should we consider this function? Note that both $x$ and $y$ must be positive; how does the constraint that girth plus length is 108 inches produce intervals of possible values for $x$ and $y\text{?}$
Find the absolute maximum of the volume of the parcel on the domain you established in (f) and hence also determine the dimensions of the box of greatest volume. Justify that you’ve found the maximum using calculus.

Hint.

Sketch a picture of a box with a square end; you’ll want to add to it as the problem progresses. Note that the question’s wording suggests the square end should be the smallest end. What does that tell you about which of $x$ and $y$ is bigger?
What are you trying to maximize or minimize?
Recall that the girth is the perimeter of the smallest end.
Pick a variable to isolate and use basic algebra to solve for it.
Substitute your answer to (d) into your formula from (b).
Each term in the sum has to be between $0$ and $108\text{.}$
Find the critical numbers of the volume function and use the first or second derivative test to ascertain which is a maximum. Once you have an answer, don’t forget to check that it actually works: the fact that the girth measures the smallest end of the box tells you which of $x$ or $y$ should be bigger (based on the formulas you used). Does that relationship hold with the dimensions you found? If not, you’ll need to rework the problem.

Answer.

Since the problem says $y$ is the length of the longer side, we assume $x\le y$ and that the square end is smallest.

Figure 3.29. A rectangular parcel with smallest end a square.
We want to maximize volume: $V=x^2y\text{.}$
If the square end is smallest, then $4x+y=108$ and $x\le y\text{.}$
$y=108-4x\text{.}$
$V(x)=x^2(108-4x)\text{.}$
$0\lt x\le 21.6\text{,}$ and $x\le y\lt 108\text{.}$
The maximum volume is $11,664\text{ in}^3$ when the box is $18''\times18''\times36''\text{.}$

Solution.

Since the problem says $y$ is the length of the longer side, we assume $x\le y$ and that the square end is smallest. However, we recognize that this may not end up being the case, and if necessary are prepared to rework the problem with the square end as the largest side.

Figure 3.30. A rectangular parcel with smallest end a square.
We are asked to maximize the volume of the box. Since the box has a square end with side length $x$ inches and is $y$ inches long, the volume formula is

\begin{equation*} V=x^2y\text{,} \end{equation*}

measured in cubic inches.
We are working under the assumption that the square end is the smallest end, which means the perimeter of the square face will be the girth and the other side will be the length. This yields the relation

\begin{equation*} 4x+y=108\text{.} \end{equation*}

We note again that we have the assumed constraint $x\le y\text{.}$
Since $4x+y=108\text{,}$ we can say that $y=108-4x\text{.}$
Since $V=x^2y$ and $y=108-4x\text{,}$ an alternate expression for the parcel’s volume is

\begin{equation*} V(x)=x^2(108-4x)=108x^2-4x^3\text{.} \end{equation*}
We know that $x$ and $y$ must both be positive, so we have $0\lt x,y\text{.}$ Additionally, because the girth is given by $4x$ rather than by $2x+2y$ (the perimeter of the other sides), we can say that $x\le y\text{.}$ Finally, since $4x+y=108$ and $x\le y\text{,}$ it must be the case that $5x\le4x+y=108\text{,}$ so $x\le 108/5=21.6\text{.}$ Likewise, since $x$ is positive, we must have $y\lt 108\text{.}$ In sum, we have the intervals:

\begin{equation*} 0\lt x\le21.6 \text{ and }x\le y\lt108\text{.} \end{equation*}
We now need to maximize the function $V(x)=108x^2-4x^3$ on the interval $(0,21.6]\text{.}$ We begin by differentiating $V\text{,}$ finding

\begin{equation*} V'(x)=216x-12x^2=12x(18-x)\text{.} \end{equation*}

As $V'(x)$ is a polynomial, it is defined everywhere; to find the critical numbers of $V\text{,}$ then, we need only solve the equation $V'(x)=0\text{.}$ Doing so yields the critical numbers $x=0$ and $x=18\text{.}$ On the interval $0\lt x\le21.6\text{,}$ the only critical number is $18\text{.}$

Testing to check that this is a maximum, the first derivative test says $V'(x)\gt0$ for $0\lt x\lt 18$ and $V'(x)\lt0$ for $18\lt x\text{.}$ Thus $V(18)$ is a relative maximum and, on the interval $(0,21.6]$ (or more generally, the interval $(0,\infty)$), an absolute maximum. Further supporting this conclusion is the second derivative test, where $V''(x)=216-24x$ is zero only when $x=9\text{,}$ and $V''(18)\lt0\text{,}$ showing that $V$ is concave down on (at least) the interval $(9,21.6)\text{.}$

We have firmly established that the maximum volume occurs when $x=18\text{.}$ This coincides with the length $y=108-4(18)=36$ and the volume $V=(18)^2(36)=11,664\text{.}$ Therefore the maximum volume of a parcel with a square end (provided that end is the smallest end of the box
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It turns out that when we assume the square end is the larger end, we end up with the dimensions $24\times24\times18$ and a volume of $10,368\text{ in}^3\text{,}$ so our process did indeed find the maximum volume of a parcel with a square end.
) subject to the dimension constraints of the USPS is $11,664$ cubic inches. This volume is achieved when the package is $18$ inches by $18$ inches by $36$ inches.

Subsection 3.3.1 More Applied Optimization Problems

Many of the steps in Example 3.27 are ones that we will execute in any applied optimization problem. We briefly summarize those steps here to provide an overview of our approach in subsequent questions:

Draw a picture and introduce variables.

It is essential to first understand what quantities are allowed to vary in the problem and then to represent those values with variables. Drawing a picture with these variables labeled is a particularly helpful way to accomplish this; in some cases, drawing several pictures might be useful. A nice example of this can be seen at http://gvsu.edu/s/99
¹⁷
gvsu.edu/s/99
, where the choice of where to bend a piece of wire into the shape of a rectangle determines both the rectangle’s shape and area.
Identify the quantity to be optimized as well as any key relationships among the variable quantities.

Essentially this step involves writing equations that involve the variables that have been introduced: one equation will represent the quantity whose minimum or maximum is sought; there may be additional equations showing how multiple variables in the problem are interrelated.
Determine a function of a single variable that models the quantity to be optimized.

It may be necessary to eliminate one or more variables in the function formula using the secondary equations that were found in the previous step. For example, in Example 3.27 we initially found that $V = x^2 y\text{;}$ the additional relationship that $4x + y = 108$ allowed us to substitute $108-4x$ for $y$ in the volume equation, thus expressing the volume as a function of the single variable $x\text{.}$
Decide the domain on which to consider the function being optimized.

Often the physical constraints of the problem will limit the possible values that the independent variable can take on. Thinking back to the picture that describes the overall situation and any relationships among variables in the problem often helps identify the smallest and largest values of the input variable.
Use calculus to identify the absolute maximum and/or minimum of the quantity being optimized.

This always involves finding the critical numbers of the function first. Then depending on the domain, we either construct a first derivative sign chart (for an open or unbounded interval) or evaluate the function at the endpoints and critical numbers (for a closed, bounded interval).
Finally, we make certain we have answered the question that was posed.

Does the question seek the absolute maximum of a quantity, or do we need the values of the variables that produce the maximum? In other words, finding the maximum volume of a parcel is different from finding the dimensions of the parcel that yield that volume.

The remaining examples involve a variety of quantities to be optimized. Each task presents an opportunity to practice the above steps, possibly with some adaptations to suit the context of the problem.

Example 3.31.

A soup can in the shape of a right circular cylinder is to be made from two materials. The material for the side of the can costs $0.015 per square inch and the material for the lids costs $0.027 per square inch. Suppose that we desire to construct a can that has a volume of 16 cubic inches. What dimensions minimize the cost of the can?

Draw a picture of the can and label its dimensions with appropriate variables.
Use your variables to determine expressions for the volume, surface area, and cost of the can.
Determine the total cost function as a function of a single variable. What is the domain on which you should consider this function?
Find the absolute minimum cost and the dimensions that produce this value.

Hint.

Note that both the radius and the height of the can are variable.
Remember that volume is the area of the base times the height, while surface area can be thought of in terms of the area of the two lids, plus the area of the “side” of the can.
Use the fact that $V = 16$ to write one of the variables in terms of the other.
Differentiate the total cost function and find its critical number(s) first.

Answer.

Let the can have radius $r$ and height $h\text{.}$
The volume is $V = \pi r^2 h\text{;}$ the surface area is $S = 2 \pi r^2 + 2 \pi r h\text{;}$ the total cost is $C = 2 \pi r^2 \cdot 0.027 + 2 \pi r h \cdot 0.015\text{.}$
The cost as a single-variable function is $C(r) = 0.054 \pi r^2 + 0.48 \frac{1}{r}\text{,}$ considered on the interval $r \gt 0\text{.}$
The minimizing radius is $r = \sqrt[3]{ \frac{0.48}{0.108 \pi} } \approx 1.123$ inches; the minimizing height is $h \approx 4.041$ inches; the minimum cost is $C(1.12259) \approx \$0.642\text{.}$

Solution.

We let $r$ be the radius of the base of the cylindrical can and $h$ be its height.
The volume of a cylinder is the area of the base times the height, so $V = \pi r^2 h\text{.}$ Surface area is the area of the lids plus the area of the “side”; we can think of the side as a rectangle (if we were to cut it and unroll it) with height $h$ and width $2\pi r\text{,}$ the perimeter of the base. Hence the surface area of the can is

\begin{equation*} S = 2 \pi r^2 + 2 \pi r h\text{.} \end{equation*}

Finally, the total cost (in dollars) is the cost of the lids plus the cost of the sides, which is

\begin{equation*} C = 2 \pi r^2 \cdot 0.027 + 2 \pi r h \cdot 0.015\text{.} \end{equation*}
Because the volume is fixed at 16 cubic inches, we know that $16 = \pi r^2 h\text{.}$ Choosing to solve for $h\text{,}$ we find that $h = \frac{16}{\pi r^2}\text{.}$ We can then substitute this expression into the total cost formula in place of $h\text{,}$ yielding

\begin{align*} C(r) =\mathstrut\amp 2 \pi r^2 \cdot 0.027 + 2 \pi r \left( \frac{16}{\pi r^2} \right) \cdot 0.015\\ =\mathstrut\amp 0.054 \pi r^2 + \frac{0.48}{r}\text{.} \end{align*}

We note that the only constraint on $r$ is that $r \gt 0\text{,}$ hence this is the domain on which we seek to minimize $C\text{.}$
To differentiate the cost function, we first recall that $\frac1r=r^{-1}$ and then find

\begin{align*} C'(r)=\mathstrut\amp 0.108\pi r-0.48r^{-2}\\ =\mathstrut\amp 0.108 \pi r - \frac{0.48}{r^2}\text{.} \end{align*}

Note that $C'(r)$ is undefined at $r=0\text{,}$ but this point is not in the domain of $C\text{,}$
¹⁸
Here, we mean both that $C(0)$ is undefined and that contextually we require $r\gt0$ as discussed in (c). Because $C(0)$ is undefined, $0$ is not a true critical number of $C\text{;}$ however, it is still a value at which the sign of $C'$ can change, and we can only disregard it here because the contextual supposition that $r\gt0$ stipulates that we need only find critical numbers on the interval $(0,\infty)\text{.}$
so we only need to consider critical numbers at which the derivative is zero. Hence we set $C'(r) = 0$ and solve for $r\text{,}$ finding that

\begin{equation*} 0.108 \pi r = \frac{0.48}{r^2}\text{,} \end{equation*}

so that $r^3 = \frac{0.48}{0.108 \pi} \approx 1.415\text{,}$ from which it follows that $r = \sqrt[3]{ \frac{0.48}{0.108 \pi} } \approx 1.123$ is the only critical number of $C\text{.}$ At this point we can use either the first or second derivative test to justify that $C$ has an absolute minimum at $r = \sqrt[3]{ \frac{0.48}{0.108 \pi} }\text{.}$ We choose to use the second derivative test; note that $C''(r) = 0.108 \pi + 0.96 \frac{1}{r^3}\text{,}$ which is always positive for $r \gt 0\text{.}$ Thus $C$ is always concave up on the relevant domain ($r \gt 0$), and consequently $r = \sqrt[3]{ \frac{0.48}{0.108 \pi} } \approx 1.123$ is where the absolute minimum of $C$ occurs. In addition, we note that since $h = \frac{16}{\pi r^2}\text{,}$ the corresponding $h$ value is $h \approx 4.041\text{,}$ and the overall minimum value is $C\left(\sqrt[3]{ \frac{0.48}{0.108 \pi} }\right) \approx 0.641\text{.}$

Thus a can with radius approximately $1.123$ inches and height approximately $4.041$ inches will be cheapest among all cans with a volume of $16$ cubic inches. The cost of constructing a can with these dimensions is about $\$0.641\text{,}$ or roughly $64$ cents.

Familiarity with common geometric formulas is particularly helpful in problems such as the one in Example 3.31. Sometimes those involve perimeter, area, volume, or surface area. At other times, the constraints of a problem introduce right triangles (where the Pythagorean Theorem applies) or other functions whose formulas provide relationships among the variables.

Example 3.32.

A hiker starting at a point $P$ on a straight road walks east towards point $Q\text{,}$ which is on the road and three kilometers from point $P\text{.}$ Two kilometers due north of point $Q$ is a cabin. The hiker will walk down the road for a while, at a pace of $8$ kilometers per hour. At some point $Z$ between $P$ and $Q\text{,}$ the hiker leaves the road and makes a straight line towards the cabin through the woods, hiking at a pace of $3$ kph, as pictured below in Figure 3.33. In order to minimize the time to go from $P$ to $Z$ to the cabin, where should the hiker turn into the forest?

Figure 3.33. A hiker walks from $P$ to $Z$ to the cabin, as pictured.

Hint.

Let $x$ be the distance from $Z$ to $Q\text{.}$ What is the distance from $P$ to $Z$ in terms of $x\text{?}$ How about the distance from $Z$ to the cabin? How does time depend on distance and walking speed?

Answer.

The absolute minimum time the hiker can achieve is about $0.993$ hours, which is attained by hiking about $2.2$ km from $P$ towards $Q$ and then turning into the woods for the remainder of the trip.

Solution.

We begin by letting $x$ be the distance from $Z$ to $Q\text{.}$ Since it is $3$ km from $P$ to $Q\text{,}$ the distance from $P$ to $Z$ is $3-x$ km. Furthermore, we can use the Pythagorean Theorem to find that the distance from $Z$ to the cabin is $\sqrt{4+x^2}$ km.

Next we want to determine the hiker’s time as a function of $x\text{.}$ Because distance equals rate times time, it follows that time is distance divided by rate. The hiker travels along the road for $3-x$ km at a pace of $8$ km/hr, thus her time on the road is

\begin{equation*} T_r = \frac{3-x}{8} \text{ hr}\text{.} \end{equation*}

Once she enters the woods, her pace drops to $3$ km/hr and she continues for $\sqrt{4+x^2}$ km before reaching the cabin. She spends a total of

\begin{equation*} T_w = \frac{\sqrt{4+x^2}}{3} \end{equation*}

hours hiking in the woods.

Altogether, the time it takes the hiker to reach the cabin is given by the function

\begin{equation*} T(x) = \frac{3-x}{8} + \frac{\sqrt{4+x^2}}{3}\text{.} \end{equation*}

Because the only values of $x$ that make sense to use are $0 \le x \le 3$ (since walking away from or past $Q$ before turning to the cabin both add unnecessary time to the trip), we use this domain for $T$ and now seek the absolute minimum of $T$ on $[0,3]\text{.}$ We find that

\begin{equation*} T'(x) = -\frac{1}{8} + \frac{1}{3} \cdot \frac{1}{2} \left(4+x^2\right)^{-\frac{1}{2}} (2x) = -\frac{1}{8} + \frac{x}{3\sqrt{4+x^2}}\text{.} \end{equation*}

Note that since $4+x^2\ge2$ for every value of $x\text{,}$ $T'$ is defined everywhere. Setting $T'(x) = 0$ and solving for $x\text{,}$ we have $\frac{x}{3\sqrt{4+x^2}} = \frac{1}{8}\text{,}$ so $8x = 3\sqrt{4+x^2}\text{.}$ Squaring both sides yields

\begin{equation*} 64x^2 = 9(4+x^2) = 36 + 9x^2\text{.} \end{equation*}

Hence $55x^2 = 36\text{,}$ so $x = \sqrt{\frac{36}{55}} \approx 0.809\text{.}$ (We don’t consider the critical number $x = -\sqrt{\frac{36}{55}}$ because this doesn’t lie in the relevant domain of $T\text{.}$)

Finally, we evaluate $T$ at the only critical number in the interval and at the interval’s endpoints. Doing so, we find $T(0) = \frac{3}{8} + \frac{2}{3} \approx 1.0417\text{,}$ $T(3) = \frac{\sqrt{13}}{3} \approx 1.202\text{,}$ and $T\left(\sqrt{\frac{36}{55}}\right) = \frac{3}{8} + \frac{\sqrt{55}}{12} \approx 0.993\text{.}$ Thus the absolute minimum time the hiker can achieve is about $0.993$ hours, which is attained by hiking about $2.2$ km from $P$ towards $Q$ and then turning into the woods for the remainder of the trip.

In more geometric problems, we often use curves or functions to provide natural constraints. For instance, we could investigate which isosceles triangle that circumscribes a unit circle has the smallest area

¹⁹

You can explore for this problem for yourself at http://gvsu.edu/s/9b.

, or we might seek the rectangle of largest area that fits within a region bounded by a parabola

²⁰

An example of this is shown at http://gvsu.edu/s/9c.

. The next example is similar to the latter problem.

Example 3.34.

Consider the region in the $xy$-plane that is bounded by the $x$-axis and the function $f(x) = 25-x^2\text{.}$ Construct a rectangle whose base lies on the $x$-axis and is centered at the origin, and whose sides extend vertically until they intersect the curve $y = 25-x^2\text{.}$ Which such rectangle has the maximum possible area? Which such rectangle has the greatest perimeter? Which has the greatest combined perimeter and area? (Challenge: answer the same questions in terms of positive parameters $a$ and $b$ for the function $f(x) = b-ax^2\text{.}$)

Hint.

Start with a sketch of the region and a rectangle within it that satisfies the description. Let $x$ represent half the width of the rectangle’s base. How does the rectangle’s height depend on $x\text{?}$

Answer.

The maximum area is $\frac{500}{3\sqrt3}\approx96.225$ square units when the rectangle’s base is $\frac{10}{\sqrt3}$ units long. The maximum perimeter is $52$ units when the rectangle’s base is $2$ units long. The maximum combined perimeter and area is about $141.858\text{,}$ occurring when the rectangle’s base is $\frac{2\sqrt{82}-2}3\approx5.370$ units long.

Solution.

We start by drawing a picture of the region described, along with a rectangle satisfying the description in the problem. Shown below in Figure 3.35, this picture will be a useful reference throughout the problem.

Figure 3.35. The region in the $xy$-plane bounded below by the $x$-axis and above by the graph of $y=25-x^2\text{.}$ Within this region is a rectangle whose base lies on the $x$-axis and is centered at the origin, and whose top corners lie on the curve. Note that if $x$ represents half the width of the rectangle’s base, then the rectangle’s height is $f(x)=25-x^2\text{.}$

As shown in Figure 3.35, the area of the rectangle is

\begin{equation*} A(x) = (2x)(25-x^2) = 50x - 2x^3\text{.} \end{equation*}

Based on the region and our definition of $x\text{,}$ the only possible values of $x$ are for $0 \le x \le 5\text{;}$ moreover, it is evident that for either $x = 0$ or $x = 5\text{,}$ the area of the corresponding rectangle is zero. Thus the endpoints of the domain $[0,5]$ both minimize the area. Differentiating $A$ with respect to $x$ yields

\begin{equation*} A'(x) = 50 - 6x^2\text{,} \end{equation*}

so we find the critical numbers of $A$ by solving $6x^2 = 50\text{.}$ This yields $x = \pm\frac{5}{\sqrt{3}} \approx \pm2.8868\text{,}$ but we need only the positive value in this context. Taking $x=\frac5{\sqrt3}$ gives the maximum possible area of

\begin{equation*} A\left(\frac{5}{\sqrt{3}}\right) = \frac{500}{3\sqrt3} \approx 96.225 \end{equation*}

square units.

We next want to maximize the perimeter. We can see in Figure 3.35 that the rectangle’s perimeter is

\begin{equation*} P(x) = 2(2x) + 2(25-x^2)= -2x^2 + 4x + 50\text{.} \end{equation*}

It is straightforward to show that the only critical number of the quadratic function $P$ occurs when $x = 1$ and that the corresponding (maximum) perimeter is $P(1) = 52$ units.

Finally, to consider combined area and perimeter, examine the function $C(x) = A(x) + P(x)$ on the interval $0 \le x \le 5\text{.}$ You should find that the only relevant critical number is $x = \frac{\sqrt{82}-1}{3}$ and that the absolute maximum of $C$ occurs at that value.

To summarize, the rectangle with the largest area has vertices at $\left(0,\pm\frac5{\sqrt3}\right)$ and $\left(\frac{50}3,\pm\frac5{\sqrt3}\right)\text{,}$ and has an area of $\frac{500}{3\sqrt3}\approx96.225$ square units. The rectangle with the largest perimeter has vertices at $\left(0,\pm1\right)$ and $\left(24,\pm1\right)\text{,}$ and has a perimeter of $52$ units. The rectangle with the largest combined total of perimeter and area has vertices at $\left(0,\pm\frac{\sqrt{82}-1}3\right)$ and $\left(\frac{142+2\sqrt{82}}9,\pm\frac{\sqrt{82}-1}3\right)\text{,}$ and the combined total is approximately $141.858\text{.}$

Example 3.36.

A trough is being constructed by bending a $4 \times 24$ (measured in feet) rectangular piece of sheet metal.

Two symmetric folds $2$ feet apart will be made parallel to the longest side of the rectangle so that the trough has cross-sections in the shape of a trapezoid, as pictured below in Figure 3.37. At what angle should the folds be made to produce the trough of maximum volume?

Figure 3.37. A cross-section of the trough formed by folding each side to an angle of $\theta\text{.}$ The base of the trough is two feet across; the angled sides are each a foot long.

Hint.

How tall is the trough in terms of $\theta\text{?}$ How wide is the top of the trough, also in terms of $\theta\text{?}$ Find (or add) some useful right triangles to the given diagram to complete the picture. Also, is there a slightly simpler quantity that will be maximized at the same time as the volume?

Answer.

The volume is maximized when $\theta\approx1.196$ radians.

Solution.

Once we choose the angle $\theta\text{,}$ the two right triangles in the trapezoid are determined and each has a horizontal leg of length $\cos(\theta)$ and a vertical leg of length $\sin(\theta)\text{.}$ Thus the sum of the areas of the two triangles is $\cos(\theta) \sin(\theta)\text{,}$ the area of the rectangle between them is $2\sin(\theta)\text{,}$ and the total area of the trapezoidal cross-section is

\begin{equation*} A(\theta) = \cos(\theta) \sin(\theta) + 2 \sin(\theta)\text{.} \end{equation*}

Because the length of the trough is constant, the trough’s volume will be maximized by maximizing cross-sectional area. Note, too, that the domain for $\theta$ is $0 \le \theta \le \frac{\pi}{2}\text{.}$

Differentiating the area function, we find that

\begin{equation*} A'(\theta) = -\sin^2(\theta) + \cos^2(\theta) + 2 \cos(\theta)\text{.} \end{equation*}

Using the identity $\sin^2(\theta) = 1 - \cos^2(\theta)\text{,}$ it follows that

\begin{align*} A'(\theta) =\mathstrut\amp \cos^2(\theta) - 1 + \cos^2(\theta) + 2 \cos(\theta)\\ =\mathstrut\amp 2\cos^2(\theta) + 2 \cos(\theta) - 1\text{.} \end{align*}

This most recent equation is quadratic in $\cos(\theta)\text{,}$ so we can solve the equation $A'(\theta) = 0$ by letting $u=\cos(\theta)$ and first solving $2u^2 + 2u - 1 = 0\text{.}$

²¹

Note that since $0\le\theta\le\frac{\pi}2\text{,}$ we have $0\le u\le1\text{.}$

Doing so yields

\begin{equation*} u = \frac{-1 \pm \sqrt{3}}{2} \approx 0.366, -1.366\text{,} \end{equation*}

so only $u = \frac{-1 + \sqrt{3}}{2}$ might be a value of the cosine function from an angle that lies in the interval $[0,\frac{\pi}{2}]\text{.}$ To find the corresponding critical number $\theta\text{,}$ we solve $\cos(\theta) = \frac{-1 + \sqrt{3}}{2}\text{,}$ which implies $\theta = \arccos\left(\frac{-1 + \sqrt{3}}{2}\right) \approx 1.196$ radians, or $\theta \approx 68.529^\circ\text{.}$

Finally, to confirm that $A$ has an absolute maximum at $\theta = \arccos\left(\frac{-1 + \sqrt{3}}{2}\right)\text{,}$ we evaluate $A$ at this value and at the interval endpoints to find that $A(0) = 0\text{,}$ $A(\frac{\pi}{2}) = 2\text{,}$ and $A(1.196) \approx 2.202$ ft$^2\text{,}$ which is the absolute maximum possible cross-sectional area.

Therefore the volume of the trough is maximized (at around $52.844$ cubic feet) when the sides are folded at an angle of approximately $1.196$ radians.

Subsection 3.3.2 Summary

While there is no single algorithm that works in every situation where optimization is used, the following steps are helpful in most of the problems we consider: draw a picture and introduce variables; identify the quantity to be optimized and find relationships among the variables; determine a function of a single variable that models the quantity to be optimized; decide the domain on which to consider the function being optimized; use calculus to identify the absolute maximum and/or minimum of the quantity being optimized; check that the question that was asked has indeed been answered.

Exercises 3.3.3 Exercises

1. Maximizing the volume of a box.

An open box is to be made out of a 10-inch by 14-inch piece of cardboard by cutting out squares of equal size from the four corners and bending up the sides. Find the dimensions of the resulting box that has the largest volume.

Dimensions of the bottom of the box: x

Height of the box:

2. Minimizing the cost of a container.

A rectangular storage container with an open top is to have a volume of 24 cubic meters. The length of its base is twice the width. Material for the base costs 14 dollars per square meter. Material for the sides costs 9 dollars per square meter. Find the cost of materials for the cheapest such container.

Total cost = (Round to the nearest penny and include monetary units. For example, if your answer is 1.095, enter $1.10 including the dollar sign and second decimal place.)

3. Maximizing area contained by a fence.

An ostrich farmer wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle (see the figure below). There are 630 feet of fencing available to complete the job. What is the largest possible total area of the four pens?

\begin{equation*} \begin{array}{|c|c|c|c|} \hline \amp \amp \amp \\ \hline \end{array} \end{equation*}

Largest area = (include units

²²

/webwork2_files/helpFiles/Units.html

)

4. Minimizing the area of a poster.

The top and bottom margins of a poster are 8 cm and the side margins are each 2 cm. If the area of printed material on the poster is fixed at 388 square centimeters, find the dimensions of the poster with the smallest area.

\begin{equation*} \begin{array}{|c|c|c|} \hline \amp \amp \\ \hline \amp printed \amp \\ \amp material \amp \\ \hline \amp \amp \\ \hline \end{array} \end{equation*}

Width = (include units

²³

/webwork2_files/helpFiles/Units.html

)

Height = (include units

²⁴

/webwork2_files/helpFiles/Units.html

)

5. Maximizing the area of a rectangle.

A rectangle is inscribed with its base on the $x$-axis and its upper corners on the parabola $y= 1-x^2\text{.}$ What are the dimensions of such a rectangle with the greatest possible area?

Width =

Height =

6. Maximizing the volume of a closed box.

A rectangular box with a square bottom and closed top is to be made from two materials. The material for the side 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? Justify your answer completely using calculus.

7. Maximizing pasture area with limited fencing.

A farmer wants to start raising 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 fence inside the large rectangle that are parallel to two existing sides. She has decided to purchase 7500 ft of fencing. What is the maximum possible area that each of the four pens will enclose?

8. Minimizing cable length.

Two vertical poles of heights 60 ft and 80 ft stand on level ground, with their bases 100 ft apart. A cable that is stretched from the top of one pole to some point on the ground between the poles, and then to the top of the other pole. What is the minimum possible length of cable required? Justify your answer completely using calculus.

9. Minimizing construction costs.

A company is designing propane tanks that are cylindrical with hemispherical ends. Assume that the company wants tanks that will hold 1000 cubic feet of gas, and that the ends are more expensive to make, costing $5 per square foot, while the cylindrical barrel between the ends costs $2 per square foot. Use calculus to determine the minimum cost to construct such a tank.

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