5.10 Introduction to Optimization Problems

TLDR

Optimization problems ask you to find the largest or smallest value of a quantity, like the most volume or the least surface area. You set up an objective function, use a constraint to reduce it to one variable, then use derivatives to find and confirm the minimum or maximum. This is the AP Calculus skill of using the derivative to locate extreme values in real contexts.

Why This Matters for the AP Calculus Exam

Optimization pulls together everything from earlier in Unit 5: critical points, the First Derivative Test, the Second Derivative Test, and checking endpoints. On the AP Calculus exam, you may see optimization in both multiple-choice and free-response settings, where you have to set up a function, find where its derivative is zero or undefined, and justify that you found a true maximum or minimum. Clear setup and justification matter for full, readable work, so naming your variables, stating the constraint, and confirming the type of extremum all help.

Key Takeaways

• Optimization means finding a minimum or maximum value of a quantity using the derivative.
• Write the quantity you want to optimize as an objective function, often with two variables at first.
• Use the constraint equation to rewrite one variable in terms of the other so you have a single-variable function.
• Find critical points where the derivative equals zero or does not exist.
• Confirm whether each critical point is a minimum or maximum using the First or Second Derivative Test, and check endpoints when the interval is closed.
• Answer the actual question asked, including the value or dimensions, not just the x value of the critical point.

What Optimization Problems Are

Optimization problems involve finding the best possible solution, usually the largest or smallest value of some quantity. "Optimizing" means maximizing or minimizing.

Calculus handles this because you already know how to find the minimum or maximum of a function: take its derivative and apply either the First Derivative Test or the Second Derivative Test.

The tricky part is that these problems usually start with two variables. The fix is one extra step: find a relationship (a constraint) between the two variables, then substitute so you can "get rid of" one. That leaves a single-variable function you already know how to differentiate.

Worked Example

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

There are two variables, $x$ and $y$, and you can only differentiate with respect to one. Rewrite one variable in terms of the other.

Because $xy = -36$, you can rewrite $y$ as $-\frac{36}{x}$.

So you have:

$s(x) = x^2 + (-\frac{36}{x})^2=x^2+\frac{1296}{x^2}=x^2+1296x^{-2}$

Now find the minimum of this function using the Second Derivative Test.

First, find the critical points where $s'(x)=0$:

$s'(x)=2x-2592x^{-3}$

$2x-2592x^{-3}=0$

$2x=2592x^{-3}$

$2x\cdot x^3=2592x^{-3}\cdot x^3$

$2x^4=2592$

$x^4=1296$

$x=-6, 6$

Next, check the concavity of $s(x)$ at these points to verify they are minimums.

$s''(x)=2+7776x^{-4}$

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 $6$ minimize $s(x)$. If $x=-6$, then $y=6$. If $x=6$, then $y=-6$.

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

How to Use This on the AP Calculus Exam

Problem Solving

A reliable order of steps:

1. Identify the quantity to optimize and write it as an objective function.
2. Find the constraint equation that links your variables.
3. Substitute so the objective function depends on one variable.
4. Find critical points where the derivative equals zero or is undefined.
5. Confirm minimum or maximum using the First or Second Derivative Test, and check endpoints if the interval is closed.
6. Return to the original question and report the value or dimensions asked for.

Free Response

When you justify, refer to the actual function by name and state your reasoning. For example, "since $s''(x)>0$, $s$ is concave up, so this critical point gives a minimum." Vague statements like "it goes up" do not show calculus reasoning.

Common Trap

Stopping after you find the critical $x$ value. The question often wants the minimum surface area, the dimensions, or a $(x, y)$ pair, so plug back in and answer what was asked.

Practice Problems

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?

Question 2

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?

Answers and Solutions

Question 1

The answer is 48 square feet. Here's why:

Let the side length of the square base be $x$ and the height be $h$.

The volume can be expressed as:

$V=x^2h$

Since the volume is 32 cubic feet:

$x^2h=32$

Rewrite $h$ in terms of $x$:

$h=\frac{32}{x^2}$

You want to minimize the surface area of the open-topped play area:

$S(x)=x^2+4xh=x^2+4x\cdot (\frac{32}{x^2})$

This simplifies to:

$S(x)=x^2+\frac{128}{x}$

Find the critical points:

$S'(x)=0$

$S'(x)=2x-128x^{-2}$

$2x-128x^{-2}=0$

$2x=128x^{-2}$

$2x \cdot x^2=128x^{-2}\cdot x^2$

$2x^3=128$

$x^3=64$

$x=4$

Check the concavity to verify a minimum:

$S''(x)=2+256x^{-3}$

$S''(4)=2+256\cdot 4^{-3}=6$

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

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

The minimum surface area is:

$S=4^2+\frac{128}{4}=48 ft^2$

Question 2

The answer is 13 inches by 13 inches. Here's the work:

Let $x$ be the width of the printed text and $y$ the height. Since the area of the printed text is 121 square inches:

$y=\frac{121}{x}$

Using the one-inch margins on all four sides, the width of the card is:

$width=x+2$

And the height of the card is:

$height=\frac{121}{x}+2$

The area of the card is:

$A(x)=(x+2)(\frac{121}{x}+2)$

Expanded and simplified:

$A(x)=2x+125+\frac{242}{x}$

Find the critical points where $A'(x)=0$:

$A'(x)=2-\frac{242}{x^2}$

$2-\frac{242}{x^2}=0$

$x=-11,11$

Take the positive solution because width cannot be negative.

Verify $x=11$ is a minimum by checking concavity:

$A''(x)=\frac{484}{x^3}$

$A''(11)>0$

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

By the Second Derivative Test, $x=11$ minimizes $A(x)$.

The width of the printed area is 11 inches and the height is 11 inches.

Adding the margins, the dimensions of the card are 13 inches by 13 inches.

Common Misconceptions

• Thinking the critical $x$ value is the final answer. You usually need a value, an area, or full dimensions, so substitute back in.
• Skipping the verification step. A critical point is not automatically a minimum or maximum, so use the First or Second Derivative Test to confirm.
• Forgetting the constraint. Without using the constraint to remove a variable, you cannot reduce the problem to a single-variable function you can differentiate.
• Ignoring the domain. Negative or zero values often do not make sense for lengths, areas, or volumes, so reject solutions that break the physical setup.
• Forgetting to check endpoints on a closed interval. An absolute extremum can occur at an endpoint, not just at a critical point.
• Using vague language like "it goes up" in justifications. Refer to $f$, $f'$, or $f''$ by name and state the calculus reason.

Related AP Calculus Guides

• Unit 5 Overview: Analytical Applications of Differentiation
• 5.1 Using the Mean Value Theorem
• 5.2 Extreme Value Theorem, Global vs Local Extrema, and Critical Points
• 5.3 Determining Intervals on Which a Function is Increasing or Decreasing
• 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema
• 5.11 Solving Optimization Problems