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How to Solve Optimization Problems in Calculus

Need to solve Optimization problems in Calculus? Let’s break ’em down and develop a strategy that you can use to solve them routinely for yourself.

Overview

Optimization problems will always ask you to maximize or minimize some quantity, having described the situation using words (instead of immediately giving you a function to max/minimize). Typical phrases that indicate an Optimization problem include:

  • Find the largest ….
  • Find the minimum….
  • What dimensions will give the greatest…?
Each of these phrases should start you thinking, “I’m looking for a maximum (or minimum) value.”

Most students don’t realize that you need to complete two distinct Stages.
Before you can look for that max/min value, you first have to develop the function that you’re going to optimize. There are thus two distinct Stages to completely solve these problems—something most students don’t initially realize [Ref].

The first stage doesn’t involve Calculus at all, while by contrast the second stage is just a max/min problem that you recently learned how to solve:

Stage I. Develop the function. You must first convert the problem’s description of the situation into a function — crucially, a function that depends on only one single variable.

Stage II. Maximize or minimize that function. Now maximize or minimize the function you just developed. You’ll use your usual Calculus tools to find the critical points, determine whether each is a maximum or minimum, and so forth.

We’ll break these two big Stages into smaller steps below. To illustrate those steps, let’s together solve this classic Optimization example problem:

Example problem: Least-Expensive Closed-Top Can
solve optimation problem of can with lid
A cylindrical can, with a top lid, must contain V cm$^3$ of liquid. (A typical can of soda, for example, has V = 355 cm$^3$.) What dimensions (height and radius) will minimize the cost of metal needed to construct the can?

Stage I. Develop the function

Step 1.
In Optimization problems, always begin by sketching the situation. Always. If nothing else, this step means you’re not staring at a blank piece of paper; instead you’ve started to craft your solution.
solve optimization problem - sketch of can with radius and height labeled
The problem asks us to minimize the cost of the metal used to construct the can, so we’ve shown each piece of metal separately: the can’s circular top, cylindrical side, and circular bottom. We’ve labeled the can’s height h and its radius r. We’re looking for the values of h and r (in terms of V) that will minimize the cost of constructing the can.

Step 2.

Most frequently you’ll use your geometry knowledge.
Having drawn the picture, the next step is to write an equation for the quantity we want to optimize. Most frequently you’ll use your everyday knowledge of geometry for this step. In this problem, for instance, we want to minimize the cost of constructing the can, which means we want to use as little metal as possible. Hence we want to minimize the can’s surface area. So let’s write an equation for that total surface area:

\begin{align*}
A_\text{total} &= A_\text{top} + A_\text{cylinder} + A_\text{bottom} \\[8px] &= \pi r^2 + 2\pi r h + \pi r^2 \\[8px] &= 2\pi r^2 + 2 \pi r h
\end{align*}

That’s it; you’re done with Step 2! You’ve written an equation for the quantity you want to minimize $(A_\text{total})$ in terms of the relevant quantities (r and h).

Step 3.
Here’s a key thing to know about how to solve Optimization problems: you’ll almost always have to use detailed information given in the problem to rewrite the equation you developed in Step 2 to be in terms of one single variable.

Above, for instance, our equation for $A_\text{total}$ has two variables, r and h. We must eliminate one of them in order to proceed. The choice of which to keep and which to eliminate is arbitrary; for our solution here, we choose to keep r. (We could just as easily choose h, and develop our solution along that path instead. We’d arrive at the same final result.) Since we’re choosing to work with r, we need to use other detailed information given in the problem to write h in terms of r so we can substitute for h as a variable.


Begin subproblem.

To accomplish this substitution, we look back to see what other constraints/information the problem gave us: recall that the can must hold an amount V of liquid, where V is some number. (V might be 355 cm$^3$, for instance.) Now a cylinder of radius r and height h has a volume of $V = \pi r^2 h,$ and so we can solve for h in terms of V and constants:
$$V = \pi r^2 h$$
thus
$$h = \dfrac{V}{\pi r^2} $$
That’s our expression for h in terms of r (and the constants V and $\pi).$
End subproblem.


We can now make this substitution $h = \dfrac{V}{\pi r^2}$ into the equation we developed earlier for the can’s total area:
\[ \begin{align*}
A_\text{total} &= 2\pi r^2 + 2 \pi r h \\[8px] &= 2\pi r^2 + 2 \pi r \left( \frac{V}{\pi r^2}\right) \\[8px] &= 2\pi r^2 + 2 \cancel{\pi} \cancel{r} \left(\frac{V}{\cancel{\pi} r\cancel{^2}}\right) \\[8px] &= 2\pi r^2 + \frac{2V}{r}
\end{align*} \]

We’re done with Step 3: we now have the function in terms of a single variable, r:
$$A(r) = 2\pi r^2 + \frac{2V}{r}$$
We’re now writing $A(r)$ to emphasize that A is a function of only the single variable r, and we’ve dropped the subscript “total” from $A_\text{total}$ since we no longer need it.

This also concludes Stage I of our work: in these threes steps, we’ve developed the function we’re now going to minimize!

Notice, by the way, that so far in our solution we haven’t used any Calculus at all. That will always be the case when you solve an Optimization problem: you don’t use Calculus until you come to Stage II.

Stage II: Maximize or minimize your function

Many students don’t realize that an Optimization problem is really a max/min problem.
Many students don’t realize that an Optimization problem is really a max/min problem; it’s just one where you first have to develop the function you’re going to maximize or minimize, as we did in Stage I above. Having done that, the remaining steps are exactly the same as they are for the max/min problems you recently learned how to solve.

For instance, a few weeks ago you could have gotten this as a standard max/min homework problem:


Graph of surface area of can with top lid, versus radius
“The surface area of a can is given by $A(r) = 2\pi r^2 + \dfrac{710}{r}\,.$ Find the value of r that minimizes that area.”


You would probably automatically find the derivative $A'(r)$ (which you could equivalently write as $\dfrac{dA}{dr})$, then find the critical points, then determine whether each represents a maximum or a minimum for the function, and so forth. That’s exactly what we’re now going to do in Stage II. Hence, you already know how to do all of the following steps; the only new part to maximization problems is what we did in Stage I above.

Step 4.
We want to minimize the function
$$ A(r) = 2\pi r^2 + \frac{2V}{r}$$
and so of course we must take the derivative, and then find the critical points.

Let’s thus first find the derivative. (Time to use Calculus!) Remember that V is just a constant — it’s some number, like 355.
\[ \begin{align*}
A'(r) &= \dfrac{d}{dr}\left(2\pi r^2 + \frac{2V}{r} \right) \\[8px] &= \dfrac{d}{dr}\left(2\pi r^2 \right) + \dfrac{d}{dr}\left(\frac{2V}{r} \right) \\[8px] &= 2\pi \dfrac{d}{dr}\left(r^2 \right) + 2V \dfrac{d}{dr}\left(r^{-1} \right) \\[8px] &= 2\pi(2r) + 2V \left((-1)r^{-2} \right)\\[8px] &= 4 \pi r\, – \frac{2V}{r^2}
\end{align*} \]

The critical points occur when $A'(r) = 0$:
\[ \begin{align*}
A'(r) = 0 &= 4 \pi r\, – \frac{2V}{r^2} \\[8px] \frac{2V}{r^2} &= 4 \pi r \\[8px] \frac{2V}{4 \pi} &= r^3 \\[8px] r^3 &= \frac{V}{2\pi} \\[8px] r &= \sqrt[3]{\frac{V}{2\pi}}
\end{align*} \] We thus have only one critical point to examine, at $r = \sqrt[3]{\dfrac{V}{2\pi}}\,.$

Step 5.
Next we must justify that the critical point we’ve found represents a minimum for the can’s surface area (as opposed to a maximum, or a saddle point). We could reason physically, or use the First Derivative Test, but we think it’s easiest in this case to use the Second Derivative Test. Let’s quickly compute the second derivative, starting with the first derivative that we found above:

\[ \begin{align*}
A'(r) &= 4\pi r\, – 2V r^{-2} \\[8px] A’^\prime(r) &= 4\pi \dfrac{d}{dr}(r) -2V \dfrac{d}{dr}\left(r^{-2} \right) \\[8px] &= 4\pi -2V \left((-2)r^{-3} \right) \\[8px] &= 4\pi + \frac{4V}{r^3}
\end{align*} \]

Since $r > 0$, this second derivative $\left(A’^\prime(r) = 4\pi + \dfrac{4V}{r^3}\right)$ is always positive $\left(A’^\prime(r) > 0 \right)$. That is, the graph of A(r) versus r is always concave up. Hence this single critical point gives us a minimum (as opposed to a maximum or saddlepoint), which is what we’re after:

Can's surface area versus radius, with minimum marked

The minimum surface area occurs when $r = \sqrt[3]{\dfrac{V}{2\pi}}\,. \quad \triangleleft$

Step 6.
Now that we’ve found the critical point that corresponds to the can’s minimum surface area (thereby minimizing the cost), let’s finish answering the question: The problem asked us to find the dimensions — the radius and height — of the least-expensive can. We’ve already found the relevant radius, $r = \sqrt[3]{\dfrac{V}{2\pi}}\,.$

To find the corresponding height, recall that in the Subproblem above we found that since the can must hold a volume V of liquid, its height is related to its radius according to
$$h = \dfrac{V}{\pi r^2}\,. $$
Hence when $r = \sqrt[3]{\dfrac{V}{2\pi}}\,,$
\[ \begin{align*}
h &= \frac{V}{\pi}\,\frac{1}{r^2} \\[8px] &= \frac{V}{\pi}\,\frac{1}{\left( \sqrt[3]{\frac{V}{2\pi}}\right)^2} \\[8px] &= \frac{V}{\pi}\,\frac{2^{2/3}\pi^{2/3}}{V^{2/3}} \\[8px] &= 2^{2/3}\frac{V^{1/3}}{\pi^{1/3}} \\[8px] h &= 2^{2/3}\sqrt[3]{\frac{V}{\pi}} \quad \triangleleft
\end{align*} \] The preceding expression for h is correct, but we can gain a nice insight by noticing that
$$2^{2/3} = 2 \cdot\frac{1}{2^{1/3}}$$
and so
\[ \begin{align*}
h &= 2^{2/3}\sqrt[3]{\frac{V}{\pi}} \\[8px] &= 2 \cdot\frac{1}{2^{1/3}}\,\sqrt[3]{\frac{V}{\pi}} \\[8px] &= 2 \sqrt[3]{\frac{V}{2\pi}} = 2r
\end{align*} \] since recall that the ideal radius is $r = \sqrt[3]{\dfrac{V}{2\pi}}\,.$ Hence the ideal height h is exactly twice the ideal radius.

To summarize, we conclude that the optimum dimensions for a closed-topped can that must contain a volume V of liquid are
\[ \begin{align*}
\text{radius } r &= \sqrt[3]{\dfrac{V}{2\pi}} \quad \cmark \\[8px] \text{height } h &= 2\sqrt[3]{\dfrac{V}{2\pi}} \quad \cmark
\end{align*} \]

Step 7. One last check

You’ll lose points if you don’t answer the question that was asked.
Because Optimization solutions can be long, we recommend that before finishing you go back and check what quantity/quantities the problem requested, and make sure you’ve provided thatespecially on an exam, where you’ll lose points if you don’t answer the exact question that was asked. For example, the problem could have asked to find the value of the smallest possible surface area A, or the minimum cost.

Instead, in this case, the problem stated, “What dimensions (height and radius) will minimize the cost of metal to construct the can?” We have provided those two dimensions, and so we are done. $\checkmark$

Summary: Problem Solving Strategy

We’ve now illustrated the steps we use to solve every single Optimization problem we encounter, and they always work.

PROBLEM SOLVING STRATEGY: Optimization
The strategy consists of two Big Stages. The first does not involve Calculus at all; the second is identical to what you did for max/min problems.

Stage I: Develop the function.

Your first job is to develop a function that represents the quantity you want to optimize. It can depend on only one variable. The steps:

  1. Draw a picture of the physical situation.
    Also note any physical restrictions determined by the physical situation.
  2. Write an equation that relates the quantity you want to optimize in terms of the relevant variables.
  3. If necessary, use other given information to rewrite your equation in terms of a single variable.

Stage II: Maximize or minimize the function.

You now have a standard max/min problem to solve.

  1. Take the derivative of your equation with respect to your single variable. Then find the critical points.
  2. Determine the maxima and minima as necessary.
    Remember to check the endpoints if there are any.
  3. Justify your maxima or minima either by reasoning about the physical situation, or with the first derivative test, or with the second derivative test.
  4. Finally, check to make sure you have answered the question as asked: Re-read the problem and verify that you are providing the value(s) requested: an x or y value; or coordinates; or a maximum area; or a shortest time; whatever was asked.
Want to see how we solve other example problems?
Want to see how we use this strategy to solve other example problems? Head on over to our Optimization page for more examples with free, complete solutions.

For now, over to you:

  • What tips do you have to share about how to solve Optimization problems?
  • What questions do you have?
  • How can we make posts such as this one more useful to you?

Please comment below!

[Thanks to S. Campbell for his specific research into students’ learning of Optimization: “College Student Difficulties with Applied Optimization Problems in Introductory Calculus,” unpublished masters thesis, The University of Maine, 2013.]


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Anonymous
2 months ago

what problems can help to solve optimization

Matheno
Editor
Reply to  Anonymous
2 months ago

Thanks for asking! We have more completely solved optimization problems on this page: Optimization: Problems and Solutions.

We hope that helps!

Anonymous
5 months ago

very nicely organized! however i think it would have been more effective with some numbers, instead of variables. it can get hard to follow, especially when there’s multiple(in this case). but it was still lovely and easy to follow

Matheno
Editor
Reply to  Anonymous
5 months ago

Thank you for your nice comment, and for your suggestion. We’ll keep it in mind for future posts. For now: thanks very much!

Jon
8 months ago

I am having a tough time differentiating what it means when they say minimize(or maximize) a situation, as in how do the answers vary one from the other? I understand all the steps. I just want to be sure I pick the right “direction” when presented with an optimization problem….

Matheno
Editor
Reply to  Jon
8 months ago

Thanks for asking, Jon!

We’re happy to try to answer your question. First, we’d like to ask for a bit of clarification so we can be sure we address the question you actually have. Can you say a bit more about “how do the answers vary one from the other” ? Even better, could you give us an example, or two, of a problem where you don’t know what direction to pick initially? Is it that you’re not sure whether you need to maximize or minimize a particular quantity, or that when you’re presented with a situation you don’t know where to begin at all, or ??

And we know it’s tough to ask a clear question when you’re learning a new topic, so please just add whatever you can.

Thanks very much, and thanks again for posting your question, which I’m sure other students have as well!

Anonymous
1 year ago

very helpful

Matheno
Editor
Reply to  Anonymous
1 year ago

We’re glad you found this helpful. Thanks for letting us know!

farah
1 year ago

this was really great and perfectly organized. I appreciate the effort and the easiness of this explanation. thank you very much

Matheno
Editor
Reply to  farah
1 year ago

Thanks for your note, Farah. We’re very happy to have helped! 🙂