5.10 Introduction to Optimization Problems

We previously learned how to find the minimum or maximum value of a function on an interval using either the First Derivative Test or the Second Derivative Test. What significance does figuring out these pieces of information have? Well, they can help us solve optimization problems!

👍 Optimization Problems

You may ask, what are optimization problems? If we think about what the word “optimizing” means, we see that these types of problems involve finding the best possible solution, often looking to maximize or minimize a certain quantity.

So how can we use calculus to solve these kinds of problems? Well, if we go back to its definition, we see that we want to “look to maximize or minimize a certain quantity.” Do we know how to find the minimum or maximum value of a function? We certainly do! How? By taking the derivatives of the function (applying either the First Derivative or Second Derivative Tests)!

One tricky part about optimization problems is that they usually add one more variable that we have to account for. But don’t fear, this just means you have to do the extra step of finding a relationship between two of the variables so that you can “get rid of” one through substitution.

✏️ Optimization Walkthrough

Let’s go through a problem below to get a better idea of how to solve this type of problem:

Let $s = x^2 + y^2$. If $xy = -36$, what $x$ and $y$ values minimize $s$?

Since there are two variables, $x$ and $y$, and we do not know how to differentiate with respect to more than one variable, we should look to re-write one of them in terms of the other. This way we can differentiate with respect to just one variable, which we’ve done many times before!

Because it is given that $xy = -36$, we can re-write $y$ to be $-\frac{36}{x}$.

So we have,

Note: $s$ becomes $s(x)$ because we have re-written the function of $s$ rewritten in terms of $x$

Now, we just have to find the minimum of this function!

To do so, let’s apply the Second Derivative Test.

First, we need to find the critical points of the function. Since we know that critical points are when $s'(x)=0$:

Next, we need to determine the concavity of $s(x)$ at these points to verify they are minimum(s).

Since the power of $x$ is even, $s''(x)$ is positive for any $x$. So, $s(x)$ is concave up at both $x=-6$ and $6$.

Therefore, both $x=-6$ and $6$ will minimize $s(x).$ If $x=-6$, based on the given product of $x$ and $y$, $y=6$. Similarly, if $x=6$, $y=-6$.

The final solution is $(-6, 6)$ and $(6, -6)$. Great work!

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📝 Optimization Practice Problems

Now, let’s do some practice on your own!

❓ Optimization Problems

Optimization Question 1

An open-topped play area with a square base is designed to hold $32$ cubic feet of sand. What is the minimum exterior surface area of the play area?

Optimization Question 1

$121$ square inches of text is to be printed on a card. If there are to be exactly one-inch margins around all four sides of the text, what is the width and height of the smallest card that can be used?

✅ Optimization Answers and Solutions

Optimization Question 1

The answer to this question is $48$ square feet. Here’s why:

First, let us set the side length of the square base to be $x$ and the height of the play area to be $h$.

This means that the volume of the play area can be expressed as

Since it is given that the volume is $32$ cubic feet,

Since we can only differentiate with respect to one variable, we can re-write $h$ in terms of $x$,

In this problem, we want to minimize the surface area of the open-topped play area, which we can express as

This further simplifies to

Now, we just have to find the minimum of this function! Let’s apply the Second Derivative Test.

First, we need to find the critical points of the function.

Next, we need to determine the concavity of $S(x)$ at this point to verify it is a minimum.

Since $6>0$, $S(x)$ is concave up at $x=4$.

Therefore, by the Second Derivative Test, $x=4$ minimizes $S(x)$.

This makes the minimum surface area of the open-topped play area

Optimization Question 2

The answer to question 2 is $13$ inches by $13$ inches. Here’s the work!

If $x$ is the width of the printed text and $y$ is the height of the printed text because we are given that the area of the printed text is $121$ square inches, we can re-write $y$ as:

Using the information given about the one-inch margins on all four sides of the text, we can write the width of the card as:

And write the height of the card as:

This means that the area of the card is

This can be expanded and simplified into

Now, we just have to find the minimum of this function! Let’s apply the Second Derivative Test.

First, we need to find the critical points of the function, which are the points where $A'(x)=0$.

We want the solution that is positive because width cannot be negative.

To verify that $x=11$ is a minimum, we need to determine the concavity of $A(x)$ at this point.

$A(x)$ is concave up at $x=11$.

Therefore, by the Second Derivative Test, $x=11$ minimizes $S(x)$.

This means that the width of the printed area is $11$ inches and the height of the printed region is $11$ inches as well.

Therefore, the dimensions of the card itself are $13$ inches by $13$ inches.

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💫 Closing

Great work so far! In the next study guide, we’ll do a couple more optimization problems to really get that practice in. You’re so close to finishing unit 5 of AP Calculus. 🥳