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Differential Calculus: From Practice to Theory

Eugene Boman, Robert Rogers

Section 9.7 Optimization Problems

So far in this chapter the examples and problems we’ve looked at were chosen to emphasize different aspects of optimization problems. Our intention was to show you the subtleties involved one at a time so you could understand them separately.

Subsection 9.7.1 Optimization Strategy

Of course, this is entirely unrealistic. Real problems rarely tell you where the difficulties will lie, so in this final section we offer you a selection of problems. It is up to you to
  1. Find an appropriate objective function.
  2. Identify the constraints.
  3. Decide whether the natural domain of the problem consists closed intervals or not.
  4. Identify any POTPs. If the problem domain consists only of closed intervals use the Extreme Value Theorem 9.5.21. Otherwise use the First Derivative Test 9.5.4 or the Second Derivative Test 9.6.14.
  5. Identify which, if any, of the POTPs are global extrema.
  6. Re-read the problem and confirm that you have solved it.
Step 6 is more important than it may seem to be. When the problems are complex it is easy to lose track of the actual question and stop too soon, or to compute quantities that weren’t needed. Always make sure you have completely solved the problem stated and not some nearby problem.
Recall that in Example 3.3.4 we observed that most people can see intuitively that of all rectangles with a fixed perimeter the one that encloses the greatest area will be a square. We gave an algebraic demonstration of this in Chapter 3 but now that we know how to use Calculus to optimize let’s revisit this problem.

Example 9.7.1. Maximizing the Area of a Rectangle Redux, Redux.

Show that of all rectangles with a fixed perimeter, \(P\text{,}\) the one with the greatest area is a square.
We first solved this problem algebraically in Example 3.3.4. We solved it again analytically (that is, using Calculus) using Fermat’s Theorem 9.1.1 in Example 9.2.9. But Fermat’s Theorem only allows us to find a maximum that we already know exists. Since we did not have the tools necessary to show that we actually had a maximum, we relied on our intuition and assumed that a maximum existed. We will solve this problem one last time, showing that there must be a maximum this time.
We will follow the outline above.
  1. The objective function is
    \begin{equation} A=xy, \tag{9.26} \end{equation}
  2. The only constraint is
    \begin{equation} P=2x+2y \tag{9.27} \end{equation}
    where \(P\) is constant.
  3. Since we are talking about a length it must be that \(x\ge0\text{.}\) From equation (9.27) we see that \(x\) must be less that or equal to \(\frac{P}{2}\text{.}\) Otherwise the perimeter will be greater than \(P\text{.}\) So the domain of our problem is \(0\le x\le \frac{P}{2}\text{.}\)
    Notice that, strictly speaking, if \(x=0\) or \(x=\frac{P}{2}\) we don’t actually have a rectangle. But it is simpler to use the Extreme Value Theorem 9.5.21 than to find the increasing and decreasing intervals using the First Derivative Test 9.5.4. Since the Extreme Value Theorem requires a closed domain we make things simpler by allowing “zero width” (or height) rectangles.
  4. Since the problem domain is a closed interval we will use the Extreme Value Theorem. To find the POTPs we will need \(\dfdx{A}{x}\text{.}\) From equation (9.26) we see that
    \begin{equation} \dx{A}=x\dx{y}+y\dx{x}.\tag{9.28} \end{equation}
    From equation (9.27) we have
    \begin{equation} \dx{y}=-\dx{x}\tag{9.29} \end{equation}
    and also
    \begin{equation} y=\frac12\left(P-2x\right) \tag{9.30} \end{equation}
    Combining equation (9.27), equation (9.28), and equation (9.30) we see that
    \begin{equation*} \dfdx{A}{x}=\frac{P}{2}-2x. \end{equation*}
    The only solution of \(\dfdx{A}{x}=0\) is \(x=\frac{P}{4}\text{,}\) so the POTPs for this problem are the end points, \(x=0\) and \(x=\frac{P}{2}\text{,}\) and \(x=\frac{P}{4}\text{.}\)
  5. The area, \(A=xy\text{,}\) will be zero when \(x=0\) and when \(x=\frac{P}{2}\) (why?) so by the Extreme Value Theorem the maximum occurs when \(x=\frac{P}{4}\text{.}\)
  6. This problem asks us to show that the maximal rectangle must be a square, so we need to show that \(x=y\text{.}\) We know that the maximum occurs when \(x=\frac{P}{4}\text{.}\) Plugging this value into equation (9.30) we see that \(y\) is also equal to \(\frac{P}{4}\text{.}\) Thus the maximal rectangle is a square as stated.

Drill 9.7.2.

It is still intuitively obvious that when \(x=\frac{P}{2}\) the area in Example 9.7.1 is a maximum. We have always believed this.
But belief is not knowledge. Use the First Derivative Test 9.5.4 to show that than maximal area is attained at \(x=\frac{P}{2}\text{.}\)
In Example 9.7.1 we were careful to follow the steps we listed at the beginning of this section in the sequence given. That list is helpful. That’s why we wrote it out. However, it would be a mistake to think of it as a recipe to be followed blindly. Think of it as an outline. An outline is a guide, not a recipe. It is not necessary to follow it rigidly, step–by–step. If you find it simpler to do these steps in a different order for some problem there is no reason that you shouldn’t. But before you declare the problem solved look back at this outline and make sure that you have done everything that needs to be done. This is what an outline is for.
In Example 3.4.3 in Chapter 3 we considered the related problem of minimizing the perimeter when the area is fixed. The following problem asks you to solve this problem using Calculus. Notice that just as above we have
\begin{equation*} A=xy \end{equation*}
and
\begin{equation*} P=2x+2y \end{equation*}
but this time \(A\) is constant, not \(P\text{.}\)

Problem 9.7.3.

Out of all rectangles with a fixed area, find the dimensions of the one with the smallest perimeter.
(a)
Identify the objective function and the constraint for this problem.
(b)
Find the natural domain of this problem and show that it is not a closed interval.
(c)
Show that we have a POTP when \(x=y\text{.}\)
(d)
Show that when \(x=y\) we have a minimum by:
  1. Using the first derivative test.
  2. Using the second derivative test.
Note that both tests will require you to use the constraint to reduce the number of variables in the objective to one. Which test was easier to use?

Subsection 9.7.2 There’s More Than One Way to Optimize

In this subsection we demonstrate how different approaches to the same problem can lead to different challenges. In this case just changing the labels on the diagram makes a big difference.
Suppose the pages of a book are to have an area of \(90\) square centimeters. The side and bottom margins are to be \(1\) centimeter in width and the top margin is to be \(1/2\) centimeter in width.
Find the dimensions of the page that has the largest possible printed area, \(P_A\text{.}\)

Subsubsection First Solution:

Since it is the printed area that is to be maximized it seems natural to label the width of the printed area, \(x\text{,}\) and the height, \(y\text{,}\) as shown below.
If we call the printed area \(P_A\) then our objective function is,
\begin{equation} P_A=xy.\tag{9.31} \end{equation}
The width of the entire page will be \(x+2\text{,}\) and the height will be \(y+3/2\text{.}\) The area of each page is constrained to be \(90\) square centimeters so our constraint is,
\begin{equation} (x+2)(y+3/2)=90.\tag{9.32} \end{equation}
Differentiating equation (9.31) we see that
\begin{equation} \dx{(P_A)}=x{\color{red}{\dx{y}}}+{\color{blue}{y}}\dx{x}\tag{9.33} \end{equation}
Differentiating equation (9.32) and solving for \(\dx{y}\) gives
\begin{equation*} \dx{y}=-\left(\frac{\textcolor{blue}{y}+3/2}{x+2}\right)\dx{x}. \end{equation*}
Solving equation (9.32) for \(y\) we have
\begin{equation} y=\frac{90}{x+2}-3/2\tag{9.34} \end{equation}
Substituting \(y\) and \(\dx{y}\) into equation (9.33) we have
\begin{equation*} \dx{(P_A)} =-x\left(\frac{\textcolor{blue}{\frac{90}{x+2}-3/2}+3/2}{x+2}\right)\dx{x}+\left(\textcolor{blue}{\frac{90}{x+2}-3/2}\right)\dx{x}. \end{equation*}
Since we’ll have to set this equal to zero and solve for \(x\) to find any POTPs it is clearly in our interest to reduce this algebraically to
\begin{equation} \dfdx{(P_A)}{x} =\frac{-90x}{(x+2)^2}+\frac{90}{x+2}-3/2.\tag{9.35} \end{equation}
Adding the fractions we see that we must solve
\begin{equation*} \dfdx{(P_A)}{x}=\frac{-90x+90(x+2)-3/2(x+2)^2}{(x+2)^2}=\frac{180-3/2(x+2)^2}{(x+2)^2}=0. \end{equation*}
At this point it is tempting to expand \((x+2)^2\text{,}\) but recall that our goal is to solve for \(x\text{.}\) Multiplying through by \((x+2)^2\) gives
\begin{equation*} 0=180-\frac{3}{2}(x+2)^2 \end{equation*}
so that
\begin{equation*} (x+2)^2=120. \end{equation*}
Solving this last equation gives \(x=\pm\sqrt{120}-2\text{.}\) But the width of a page can’t be negative so a negative solution is meaningless for this problem. Thus \(x=\sqrt{120}-2\) is our one and only POTP, and we conclude that we can maximize the printed area by taking \(x=\sqrt{120}-2\approx 8.95\) centimeters.
That was a lot of work but we are not finished The problem asks us to find the dimensions (plural) of the page. To find \(y\) we use equation (9.34) to compute
\begin{equation*} y\approx\frac{90}{\sqrt{120}-2+2}-3/2 =\frac{90}{\sqrt{120}}-3/2\approx 8.08. \end{equation*}
But we’re still not finished. The problem asks for the dimensions of the page, but we’ve only found the dimensions of the printed area. The dimensions of the page are
\begin{align*} \text{Width of page} = \textcolor{blue}{\text{width of text}} + \textcolor{red}{\text{margins}}\amp{} =\textcolor{blue}{\sqrt{120} -2} +\textcolor{red}{2} \\ \amp{}=\sqrt{120} \approx 10.95 \text{ cm}\\ \text{Height of page} = \textcolor{blue}{\text{height of text}} + \textcolor{red}{\text{margins}} \amp{}=\textcolor{blue}{\frac{90}{\sqrt{120}}-3/2} + \textcolor{red}{3/2} \\ \amp{}= \frac{90}{\sqrt{120}}\approx 8.22 \text{ cm}. \end{align*}
Finally, the last item in our outline says to make sure we have done everything that needs to be done to solve the problem. Have we?
No, clearly not. We never did step 3. As a result we don’t know for certain that we have a maximum. It may be that the critical point we’ve identified is not in the problem domain. It may be that it is in the domain, but that we have a minimum instead of a maximum. Or that we have neither. The point is that we can’t know until we have found the natural domain of the objective function, and verified that there is a global maximum at \(x=\sqrt{120}-2\approx8.95\) centimeters.
We mentioned earlier that the width of the page can’t be negative so the smallest possible value of \(x\text{,}\) the left endpoint of our interval, is zero.
Drill 9.7.5.
  1. Show that the largest possible value of \(x\) is \(x=58\) and thus the natural domain for this problem is the closed interval \([0,58]\text{.}\)
  2. Show that the maximal printed area occurs when \(x=2\sqrt{30}-2\text{.}\)

Subsubsection Second Solution

There is often more than one objective function that will work for any given problem. If your first attempt does not seem to be working --- even if you are sure you have an objective that will work --- you always have the option of going back to the beginning and trying to find another valid objective function.
For our second solution we will use the same diagram, but with different labeling. As you can see in the sketch below, this time we’ve let \(x\) and \(y\) represent, respectively, the width and height of the page rather than of the printed area.
Our objective function is still the area of the printed portion of the page, but with the updated labeling this becomes
\begin{equation} P_A=(x-2)(y-3/2).\tag{9.36} \end{equation}
Differentiating this we have
\begin{equation} \dx{(P_A)}=(x-2)\dx{y}+(y-3/2)\dx{x}.\tag{9.37} \end{equation}
The constraint is still that the area of the page is \(90\) square centimeters, but this is now given by
\begin{equation} 90=xy.\tag{9.38} \end{equation}
Differentiating the constraint gives
\begin{equation*} x\dx{y}+y\dx{x}=0, \end{equation*}
from which we see that \(\dx{y}=-\frac{y}{x}\dx{x}\text{.}\) From equation (9.38) we see that \(y=\frac{90}{x}\text{.}\) Substituting these into equation (9.37) we get
\begin{align*} \dx{(P_A)}\amp =(x-2)\left(-\frac{y}{x}\right)\dx{x}+\left(\frac{90}{x}-\frac32\right)\dx{x}\\ \amp =\left[-(x-2)\left(\frac{90}{x^2}\right)+\frac{90}{x}-\frac32\right]\dx{x}\\ \dfdx{(P_A)}{x} \amp =-(x-2)\left(\frac{90}{x^2}\right)+\frac{90}{x}-\frac32 \end{align*}
Setting this equal to zero and solving for \(x\) gives
\begin{align*} -(x-2)\left(\frac{90}{x^2}\right)+\frac{90}{x}-\frac32 \amp =0\\ \textcolor{red}{-\frac{90}{x}} -\frac{180}{x^2}+\textcolor{red}{\frac{90}{x}}-\frac32\amp =0\\ \frac{180}{x^2}-\frac32 \amp =0\\ \frac23\cdot180\amp =x^2 \text{ or}\\ x\amp =\pm\sqrt{120}. \end{align*}
As before, we know that the width of a page cannot be negative so the only POTP is \(x=\sqrt{120}\text{.}\)
This time \(x\) and \(y\) represent the width and height of the page, so the width of the printed text is will be \(\sqrt{120}-2\) as before. Similarly the height of the printed text will be \(\frac{90}{\sqrt{120}}-\frac32\text{.}\)
Problem 9.7.6.
Determine the natural domain of \(P_A(x)\) and show that \(P_A(x)\) is a global maximum when \(x=\sqrt{120}\text{.}\) Do you get the same page dimensions as before?
You will find that, quite often, simplifying complex expressions as much as possible with Algebra, Geometry, or Trigonometry will lead to easier Calculus computations. Conversely less non-Calculus work often leads to more challenging differentiations. But this is not a hard–and–fast rule. Sometimes you will find that leaving the formulas you generate unsimplified gives better insight into the problem, making things a little simpler overall, even if it makes the differentiation quite onerous.
Which parts of a given problem you find easy or difficult will depend on your skill level with your tools. With experience comes the ability to balance the workload across your skills. When you realize that one approach will lead to a complication that you know you will find difficult to handle, it can be beneficial to stop and look for another approach. If you find one, stop and try again from the beginning.
Then identify where your skill is lacking and give yourself some practice with that particular skill to improve your skills overall. Review, practice, and improve your Algebra, Geometry, and Trigonometry skills as needed.
A common mistake when abandoning a first attempt is to try to preserve and reuse computations left over from your first try. But that attempt didn’t work. You need to try something new.
Problem 9.7.7.
Did you notice that in equation (9.34) we actually had \(y\) in terms of \(x\text{?}\) That suggests yet another approach to this problem.
(a)
Find the derivative of \(y\) from equation (9.34) directly and use it to solve this problem again.
(b)
Solve this problem one last time by computing the derivative of \(y\) from equation (9.36).

Subsection 9.7.3 Selected Optimization Problems

Problem 9.7.8. Variations on a Theme.

(a) Find the Pattern.
  1. Find two positive numbers whose sum is 10, and whose product is a maximum. What is the maximum?
  2. Find two positive numbers whose sum is \(8\text{,}\) and whose product is a maximum. What is the maximum?
  3. Find two positive numbers whose sum is \(K\text{,}\) and whose product is a maximum. What is the maximum?
(b) Find the Pattern.
  1. Find two positive numbers such that the sum of one of them and twice the other is 10, and whose product is a maximum. What is the maximum?
  2. Find two positive numbers such that the sum of one of them and three times the other is 10, and whose product is a maximum. What is the maximum?
  3. Find two positive numbers such that the sum of one of them and five times the other is 10, and whose product is a maximum. What is the maximum?
  4. Find two positive numbers such that the sum of one of them and \(n\) times the other is K, and whose product is a maximum. What is the maximum?
(c) Find the Pattern.
  1. Find two positive numbers whose product is \(10\text{,}\) and whose sum is a minimum. What is the minimum?
  2. Find two positive numbers whose product is \(M\text{,}\) and whose sum is a minimum. What is the minimum?
  3. Find two positive numbers whose product is \(M\text{,}\) and the sum of one number and three times the other is a minimum. What is the minimum?
  4. Find two positive numbers whose product is \(M\text{,}\) and the sum of one number and \(n\) times the other is a minimum. What is the minimum?

Problem 9.7.9. Variations on a Theme.

You have \(500\) meters of fencing with which to build an animal pen.
(a) Find the Pattern.
  1. What is the largest area you can enclose with a freestanding rectangular pen?
  2. You want to build a rectangular pen divided into two equally sized internal pens by placing an internal fence parallel to two of the external fences. What is the largest area you can enclose?
  3. You want to build a rectangular pen divided into three equally sized internal pens by placing two internal fences parallel to two of the external fences. What is the largest area you can enclose?
  4. You want to build a rectangular pen divided into \(n\) equally sized internal pens by placing \(n-1\) internal fences parallel to two of the external fences. What is the largest area you can enclose? If \(w\) is the width and \(l\) is the length of the external pen what is \(\frac{w}{l}\text{.}\)
(b) Is there a Pattern?
  1. You want to use one side of a building as one full side of your pen as shown in the sketch below. What is the largest area you can enclose? What are the dimensions of the pen?
  2. You want to use one side of a building as one partial side of your pen as shown in the sketch below. What is the largest area you can enclose? What are the dimensions of the pen?
  3. You want to build your pen as shown in the sketch below. What is the largest area you can enclose? What are the dimensions of the pen?
  4. You want to build your pen as shown in the sketch below. What is the largest area you can enclose? What are the dimensions of the pen?
  5. You want to your pen to be subdivided into three parts as shown in the sketch below. What is the largest area you can enclose? What are the dimensions of the pen?
  6. You want to build your pen as shown in the sketch below. What is the largest area you can enclose? What are the dimensions of the pen?

Problem 9.7.10.

You have a piece of wire \(100\) inches long.
(a)
Find the length of each piece of the wire that yields the least total area bounded by the pieces when:
  1. You cut the wire into two pieces and bend each piece to form a square.
  2. You cut the wire into two pieces and bend each piece to form a circle.
  3. You cut the wire into two pieces and bend one piece to form a square and the second to form a circle.
  4. You cut the wire into two pieces and bend one piece to form a rectangle with one side three times as long as the other. Then bend the second piece to form a square.
  5. You cut the wire into three pieces and form two squares and one circle.
(b)
Repeat each problem in part (a), but this time find the length of each piece of the wire that yields the most total area bounded by the pieces.

Problem 9.7.11.

Consider the side view of a person at point \(P\) looking at a \(20\) foot tall billboard sign that is \(10\) feet off of the ground as shown in the diagram below.
(a)
Find a formula for the viewing angle \(\theta\) in terms of \(x.\) Don’t forget to put in the range for possible values for \(x.\)
(b)
Find the value of \(x\) that will maximize the viewing angle \(\theta\text{.}\)

Problem 9.7.12.

The sketch below depicts water in a circular vessel with a central hub of radius \(r\text{.}\) Notice that the surface of the water is tangent to the top if the hub. Find the value of \(r\) that maximizes the blue shaded area, \(A(r)\text{,}\) in the following diagram:

Problem 9.7.13. Find the Pattern.

(a)
The strength of a rectangular wooden beam is proportional to its width times the square of its height. What is the ratio of the height to width of the strongest beam that can be cut from a cylindrical log?
(b)
The stiffness of a rectangular wooden beam is proportional to its width times the cube of its height. What is the ratio of the height to width of the stiffest beam that can be cut from a cylindrical log? Compare this to the answer in part (a).
(c)
Suppose we have an objective function which is proportional to the width times the \(n\)th power of the height? What would the optimal ratio be now?

Problem 9.7.14.

A box with a square base and an open top must have a volume of \(32,000\) square centimeters.
Find the dimensions of the box that minimize the amount of material used to make the box.

Problem 9.7.15.

The diagram above depicts the view from above of a pole of length \(L\) being maneuvered horizontally around the \(90^\circ\) corner of two hallways of widths \(a\text{,}\) and \(b\text{.}\) Find the longest pole which can be turned around the corner.
(a)
Show that the maximum value of \(L\) is \(\left(a^\frac23+b^\frac23\right)^\frac32\text{.}\)
(b)
What would the maximum of \(L\) be in the special case where \(a=b\text{?}\) Does this answer make sense geometrically?
(c)
In the problem statement we specified that the pole was being maneuvered horizontally, meaning that the pole is kept parallel to the ground at all times. Suppose we remove that constraint and that the height of the hallway is \(h\text{.}\) What is the longest pole we can get around the corner now?

Problem 9.7.16.

(a)
A Norman window has the shape of a rectangle surmounted by a semicircle.
Find the dimensions of the window that allow the most light to enter if the perimeter of the window is \(30\) feet.
(b)
Find the dimensions of the window that allow the most light to enter if the window has the shape of a rectangle surmounted by an equilateral triangle.

Problem 9.7.17.

(a)
At which points on the curve \(y=1+40x^3-3x^5\) does the tangent line have the greatest slope?
(b)
Find the coordinates of the points on the graph of \(4x^2+y^2=4\) which are farthest from the point \((1,0)\text{.}\)

Problem 9.7.18. Find the Pattern.

(a)
Find the area of the smallest triangle formed by the \(x\) and \(y\) axes and a line passing through the point \((2,1).\)
(b)
Find the area of the smallest triangle formed by the \(x\) and \(y\) axes and a line passing through the point \((3,4).\)
(c)
Find the area of the smallest triangle formed by the \(x\) and \(y\) axes and a line passing through the point \((a,b)\text{,}\) where \(a, b\gt 0\text{.}\)

Problem 9.7.19.

(a)
We draw the right triangle whose hypotenuse starts at the origin, \((0,0)\) and ends on the Witch of Agnesi, \(W(x)=\frac{1}{1+x^2}\text{.}\) The other legs are as shown in the figure below:
Of all triangles for which \(\alpha\) is in the interval \([0,\infty)\) which one has the largest area?
(b)
Now suppose that for each function given below a triangle is constructed in the same manner used in part (a). Assume \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) and \(k\) are constants. Of all possible such triangles for which \(\alpha\) is in the interval \([0,\infty)\) which one has the largest area?
  1. \(\displaystyle f(x) = \frac{k}{x^2+1}\)
  2. \(\displaystyle f(x) = \frac{1}{x^2+3}\)
  3. \(\displaystyle f(x) = \frac{1}{5x^2+1}\)
  4. \(\displaystyle f(x) = \frac{a}{bx^2+c}\)

Problem 9.7.20. Variations on a Theme.

To prevent squirrels from eating her birdseed a mathematician wants to hang a bird feeder so that its top is \(5\) feet above the ground. To do this she attaches each end of a wire between two trees at a height of \(8\) feet above the ground, with the feeder hanging from a second wire, \(x\text{,}\) attached at the midpoint of the first as seen in the sketch below.
(a)
Find the value of \(x\) that requires the least amount of wire if the trees are \(10\) feet apart.
  1. What will the total length of the wire be?
  2. Show that \(\theta_1=\theta_2=\theta_3\text{.}\)
(b)
Find the value of \(x\) that requires the least amount of wire if the trees are \(12\) feet apart.
  1. What will the total length of the wire be?
  2. Show also that \(\theta_1=\theta_2=\theta_3\text{.}\)
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