Applied Optimization Problems

Introduction

Optimization is one of the uses of calculus in the real world. Perhaps we are a pizza parlor and wish to maximize profit. Perhaps we have a flat piece of cardboard and we need to make a box with the greatest volume. How does one go about this process?

This requires the use of maximums and minimums. We know that we find maximums and minimums via derivatives. Therefore, one can conclude that calculus will be a useful tool for maximizing or minimizing (collectively known as "optimizing") a situation.

In general, an optimization problem has a constraint that changes how we view the problem. The wording of the problem (whether subtle or not) can also drastically change how we view the problem. The constraint is the thing we will absolutely not change (such as the general shape, or the amount that we are willing to pay for the entire operation, etc.). If we can identify what we are going to change and what we are going to require stay the same, we will be on our way to solve the problem.

How to Solve

These general steps should be taken in order to complete an optimization problem.

1. Write out necessary formulas and other pieces of information given by the problem.
   • The problems should have a variable you control and a variable that you want to maximize/minimize.
   • The formulas you find may contain extra variables. Depending on how the question works out, they may be substituted out or can be ignored (which will be explained later).
2. Combine the formulas together so that the variable you want to maximize/minimize is on one side of the equation and everything else on the other.
3. Differentiate the formula. If your equation has multiple variables, pick any variable to differentiate as long as it is not the one you control (i.e. pick the variable that you could not get rid of from the formula).
   • Note that during differentiation, if you come across a variable that you have not picked, imagine it as a number and apply the necessary differentiation rule. Do not treat it as a variable in this case.
4. Set the differentiated formula to equal 0 and solve for the variable you control.
5. The value you get here is your answer. If you instead have another formula, that means that your answer depends on those other variables, which would usually be what the question asked for if you have such a situation that you have another variable to juggle to begin with.

The reason why this algorithm works comes from a few mathematical theorems which you will probably not need to know when completing these problems. Usually the problems given will be mathematically simple (in other words, there are not a lot of cases to test). However, if you wish to know, they work like this:

• A derivative of 0 is either a global or local maximum or minimum. Usually the question will tend towards answering that question without much difficulty (like always positive numbers, for example)

Examples

Volume Example

A box manufacturer desires to create a closed box with a surface area of 100 inches squared and with a square base yet rectangular sides. What is the maximum volume that can be formed by bending this material into a box?

• Write out known formulas and information

$A_{base}=x^{2}$

$A_{side}=x\cdot h$

$A_{total}=2x^{2}+4x\cdot h=100$

$V=l\cdot w\cdot h=x^{2}\cdot h$

• Write the variable $h$ in terms of $x$ in the volume equation.

$2x^{2}+4xh=100$

$x^{2}+2xh=50$

$2xh=50-x^{2}$

$h={\frac {50-x^{2}}{2x}}$

${\begin{aligned}V&=\left(x^{2}\right)\left({\frac {50-x^{2}}{2x}}\right)\\&={\frac {50x-x^{3}}{2}}\end{aligned}}$

• Find the derivative of the volume equation in order to maximize the volume

${\frac {dV}{dx}}={\frac {50-3x^{2}}{2}}$

• Set ${\frac {dV}{dx}}=0$ and solve for $x$

${\frac {50-3x^{2}}{2}}=0$

$50-3x^{2}=0$

$3x^{2}=50$

$x=\pm {\frac {\sqrt {50}}{\sqrt {3}}}$

• Plug-in the $x$ value into the volume equation and simplify

${\begin{aligned}V&={\frac {1}{2}}\left[50{\sqrt {\frac {50}{3}}}-\left({\sqrt {\frac {50}{3}}}\right)^{3}\right]\\&=68.04138174...\end{aligned}}$

Answer: $V_{max}=68.04138174...$

Volume Example II

It is desired to make an open-top box of greatest possible volume from a square piece of tin whose side is $\alpha$ by cutting equal squares out of the corners and then folding up the tin to form the sides. What should be the length of the side of a cut-out square?

If we call the side length of the cut-out squares $x$, then each side of the base of the folded box is $\alpha -2x$, and the height is $x$. Therefore, the volume function is $V(x)=x(\alpha -2x)^{2}=x(\alpha ^{2}-4\alpha x+4x^{2})=\alpha ^{2}x-4\alpha x^{2}+4x^{3}$. We must optimize the volume by taking the derivative of the volume function and setting it equal to 0. Since it does not change, $\alpha$ is treated as a constant, not a variable.

$V(x)=\alpha ^{2}x-4\alpha x^{2}+4x^{3}$

${\begin{aligned}0&=V'(x)=\alpha ^{2}-8\alpha x+12x^{2}\\&=12x^{2}-8\alpha x+\alpha ^{2}\end{aligned}}$

We can now use the quadratic formula to solve for $x$:

$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$

${\begin{aligned}x&={\frac {-(-8\alpha )\pm {\sqrt {(-8\alpha )^{2}-4(12)(\alpha ^{2})}}}{2(12)}}\\&={\frac {8\alpha \pm {\sqrt {64\alpha ^{2}-48\alpha ^{2}}}}{24}}\\&={\frac {8\alpha \pm {\sqrt {16\alpha ^{2}}}}{24}}\\&={\frac {8\alpha \pm 4\alpha }{24}}\end{aligned}}$

$x={\frac {\alpha }{6}},{\frac {\alpha }{2}}$

We reject $x={\frac {\alpha }{2}}$, since it is a minimum (it results in the base length $\alpha -2x$ being 0, making the volume 0).

Therefore, the answer is $x={\frac {\alpha }{6}}$.

Volume Example III

A cylindrical can is required to hold $1.024{\text{ L}}$ of liquid. Determine the measurements that will minimize the amount of material required to construct the can of specified volume.

This is a classic problem for optimization. The problem is we do not have any specified measurements. Luckily enough, we are given SI units, so we can write what we are given accordingly,

• Conversion factor: $1{\text{ L}}=1000{\text{ cm}}^{3}$. This means that through dimensional analysis (or basic algebra): $1.024{\text{ L}}=1024{\text{ cm}}^{3}$. Recall that this is the volume!
• Volume of a cylinder: $V=\pi r^{2}h$
• Surface area of a cylinder: $S=2\pi rh+2\pi r^{2}$

Recall what we are trying to minimize: "the amount of material required to construct the can of specified volume." This means the surface area needs to be minimized! The unfortunate part is we have two variables (it is not constant). Luckily, we can use the volume to write $h$ in terms of $r$:

$1024{\text{ cm}}^{3}=\pi r^{2}h$

$\Rightarrow {\frac {1024{\text{ cm}}^{3}}{\pi {r{\text{ cm}}}^{2}}}={\frac {1024}{\pi r^{2}}}{\text{ cm}}=h$

We can the substitute this information into the surface area formula and minimize (set the first derivative to zero):

$S(r)=2\pi r\left({\frac {1024}{\pi r^{2}}}\right)+2\pi r^{2}$. Eliminate the $\pi$ and $r$.

$\Rightarrow S(r)=2\cdot {\frac {1024}{r}}+2\pi r^{2}$ Set the first derivative equal to zero.

${\begin{aligned}\Rightarrow 0=S'(r)&={\frac {d}{dr}}\left({\frac {2048}{r}}+2\pi r^{2}\right)\\&={\frac {d}{dr}}\left({\frac {2048}{r}}\right)+{\frac {d}{dr}}\left(2\pi r^{2}\right)\\&={\frac {d}{dr}}\left(2048\cdot {r}^{-1}\right)+2\pi {\frac {d}{dr}}\left(r^{2}\right)\\&=2048\cdot {\frac {d}{dr}}\left({r}^{-1}\right)+2\pi \left(2r\right)\\&=2048\cdot (-1){r}^{-2}+4\pi r\\&=-{\frac {2048}{{r}^{2}}}+4\pi r\\&={\frac {4\pi r^{3}-2048}{{r}^{2}}}\end{aligned}}$

All that is left to do is find the critical values of the operation. This ultimately means to let the numerator equal zero and find what values of $r$ make the denominator equal to zero. Here is the operation for the numerator:

$4\pi r^{3}-2048=0$

$\Leftrightarrow 4\pi r^{3}=2048$

$\Leftrightarrow r^{3}={\frac {2048}{4\pi }}={\frac {512}{\pi }}$

$\Leftrightarrow r={\sqrt[{3}]{\frac {512}{\pi }}}={\frac {8}{\sqrt[{3}]{\pi }}}\approx 5.4623$

Here is the operation for the denominator:

${r}^{2}=0$

$r=\pm 0$

Let us note that we do not care about one of the critical values being equal to zero. This problem is asking us to find the value that makes the condition true but also minimized. While zero does minimize the amount of material used to construct the can, we cannot make a can if we do not show one. This solution would be interesting, however, if we were asked about the graph of $S(r)$. Nevertheless, our final answer may not necessarily be $r\approx 5.4623{\text{ cm}}$ and $h\approx 10.9245{\text{ cm}}$. We need to verify that the critical value we found is the correct answer. That is where we substitute values for $r\in \left(0,\,{\frac {8}{\sqrt[{3}]{\pi }}}\right)$ and $r\in \left({\frac {8}{\sqrt[{3}]{\pi }}},\,\infty \right)$ into the first derivative:

For $r\in \left(0,\,{\frac {8}{\sqrt[{3}]{\pi }}}\right)$, let $r=5$: $S'(5)={\frac {4\pi (5)^{3}-2048}{(5)^{2}}}<0$. For $r\in \left({\frac {8}{\sqrt[{3}]{\pi }}},\,\infty \right)$, let $r=6$: $S'(6)={\frac {4\pi (6)^{3}-2048}{(6)^{2}}}>0$.

Based on these results above, it is clear that $r={\frac {8}{\sqrt[{3}]{\pi }}}$ is a value that minimizes the function such that it allows us to make a can. Keep in mind, if you do not show that the value we found is a minimum, then this value could also be a maximum. Imagine if that were the case, and we suggested the company to produce a can that would maximize the amount of material used. Potentially millions of dollars of material costs would offset the profit from the optimum. Either way, we have found our answers:

Answer: the radius is $5.4623{\text{ cm}}$ and the height is $10.9245{\text{ cm}}$.

Revenue Example

Maximizing revenue

A widget firm has a linear demand function $D(q)=-{\frac {1}{5}}q+10$ where $q$ kilograms of widgets (in hundreds) are shipped and sold to customers. At what price should the widgets be sold such that the revenue is maximized?

The revenue of a firm is calculated by multiplying the quantity sold to the price. Since $D(q)$ is the price, and $q$ is the quantity sold, $R(q)=q\cdot D(q)$. The question is asking us to find the price that will maximize the quantity. This means finding the quantity is the first goal. The final goal is the price found.

Our first goal is finding the quantity that will maximize the revenue. The revenue function is maximized if the slope of the tangent line is zero at a point and the sign of the slope around those points change sign from positive to negative. This is achieved through taking the derivative and applying the first derivative test. This is made easier when the revenue function distributes the $q$:

${\begin{aligned}R(q)&=q\cdot D(q)\\&=q\cdot \left(-{\frac {1}{5}}q+10\right)\\&=-{\frac {1}{5}}q^{2}+10q\end{aligned}}$

Let us now take the first derivative:

${\frac {d}{dq}}\left[R(q)\right]={\frac {d}{dq}}\left[-{\frac {1}{5}}q^{2}+10q\right]$

${\begin{aligned}R'(q)&={\frac {d}{dq}}\left[-{\frac {1}{5}}q^{2}\right]+{\frac {d}{dq}}\left[10q\right]\\&=-{\frac {1}{5}}{\frac {d}{dq}}\left[q^{2}\right]+10{\frac {d}{dq}}\left[q\right]\\&=-{\frac {1}{5}}\left[2q\right]+10\left[1\right]\\&=-{\frac {2}{5}}q+10\end{aligned}}$

Set the first derivative equal to zero to find the critical values:

$R'(q)=0=-{\frac {2}{5}}q+10$

$\Leftrightarrow -10=-{\frac {2}{5}}q$

$\Leftrightarrow 25=q$

Take note of the original revenue function. Because the revenue function is a quadratic that has its $q^{2}$ term multiplied by a factor less than 0, it is guaranteed there is only one critical value, and it must be a maximum. The calculus-based approach simply confirms this truth (using the first derivate test):

For any $q<25$, $R'(q)>0$. For any $q>25$, $R'(q)<0$. Therefore, the critical value $q=25$ must be a maximum.

All that is left to do is input this domain into the demand function to find the price the widgets should be sold at. This is a simple substitution exercise:

${\begin{aligned}D(25)&=-{\frac {1}{5}}(25)+10\\&=-5+10\\&=5\end{aligned}}$

The widgets should be sold at $$5$ to maximize the revenue.

Resources

• Applied Optimization PowerPoint file created by Dr. Sara Shirinkam, UTSA.

• Applied Optimization Worksheet

• Cal1 Applied Optimization Part1 video lecture by Instructor Beatty

• Cal1 Applied Optimization Part2 video lecture by Instructor Beatty

Licensing

Content obtained and/or adapted from:

• Optimization, Wikibooks: Calculus under a CC BY-SA license