AP Calculus Optimization- Practice Questions and Solutions
AP Calculus Optimization Problems: What You Actually Need to Know
Optimization problems show up on the AP exam every year. They're also the problems students lose the most points on. Not because the math is hard—it's not—but because students don't understand what they're actually supposed to do.
Here's the deal: optimization is just finding maximums and minimums. You've been doing that with derivatives since unit 3. The difference is that in optimization problems, you have to create the function yourself from a word problem.
That's it. That's the whole game.
The Process: Don't Memorize, Understand
Most textbooks give you a 5 or 6-step process to memorize. Students forget steps, mix them up, and panic during the exam. Here's what you actually need to do:
- Read the problem. Figure out what quantity you want to maximize or minimize.
- Identify your constraints. These are the "given" conditions that limit your options.
- Write the primary equation—the one containing the quantity you want to optimize.
- Use your constraints to eliminate variables until you have one equation in one variable.
- Take the derivative, set it equal to zero, and solve.
- Check that your answer makes sense (endpoints, second derivative test, or common sense).
You don't need to memorize a flowchart. You need to understand what each step accomplishes.
Common Problem Types
Fence/Enclosure Problems
These show up constantly. The setup is almost always the same: you have a fixed amount of fencing and want to enclose the maximum possible area, or you have a boundary (like a river) and want to use the least fencing.
The key insight: When you have a rectangle with a fixed perimeter, the maximum area is always a square. When you have three sides enclosed (one side is a wall or river), the optimal dimensions are always length = 2 × width.
Box Problems
Cut corners out of a sheet of cardboard and fold up the sides. Find the maximum volume. The algebra gets messy, but the process is the same: express volume as a function of one variable, take the derivative, set it to zero.
Distance Problems
Find the point on a line (or curve) closest to a given point. The distance formula is your friend here. Remember that minimizing distance squared gives you the same answer as minimizing distance, and the algebra is cleaner.
Revenue/Profit Problems
These show up in business contexts. Price × quantity = revenue. Cost functions subtract from profit. The setup is usually straightforward once you identify what changes (price or quantity) and what stays fixed.
Practice Questions with Solutions
Problem 1: The Classic Fence
A farmer has 200 feet of fencing and wants to enclose a rectangular field. One side of the field borders a river and needs no fencing. What dimensions maximize the enclosed area?
Solution:
Step 1: Let width = x (the sides perpendicular to the river) and length = y (the side parallel to the river).
Step 2: Write the constraint equation. We have 200 feet of fencing, and we're fencing three sides: 2x + y = 200
Step 3: Write the area function. A = x · y
Step 4: Substitute. From the constraint: y = 200 - 2x
So: A(x) = x(200 - 2x) = 200x - 2x²
Step 5: Differentiate and set equal to zero.
A'(x) = 200 - 4x = 0
4x = 200
x = 50
Step 6: Find y. y = 200 - 2(50) = 100
Answer: Width = 50 ft, Length = 100 ft, Maximum Area = 5,000 ft²
Notice that length is twice the width. This matches the principle I mentioned earlier.
Problem 2: The Box with No Top
A box with a square base is to be made from a 12-inch by 12-inch piece of cardboard. Square corners are cut out and the sides are folded up. What size squares should be cut to maximize volume?
Solution:
Step 1: Let x = the side length of the squares cut from each corner.
Step 2: After cutting and folding, the base dimensions are (12 - 2x) by (12 - 2x). Height = x.
Step 3: Volume function.
V(x) = x(12 - 2x)(12 - 2x) = x(12 - 2x)²
Step 4: Expand and differentiate.
V(x) = x(144 - 48x + 4x²) = 144x - 48x² + 4x³
V'(x) = 144 - 96x + 12x²
Step 5: Set equal to zero and solve.
12x² - 96x + 144 = 0
Divide by 12: x² - 8x + 12 = 0
(x - 2)(x - 6) = 0
x = 2 or x = 6
Step 6: Check both answers.
x = 6 gives dimensions of 0 × 0 × 6 = 0 volume. Eliminate.
x = 2 gives dimensions of 8 × 8 × 2 = 128 in³. This works.
Answer: Cut 2-inch squares from each corner.
Problem 3: The Point on a Line
Find the point on the line y = 2x + 3 that is closest to the origin.
Solution:
Step 1: Let the point on the line be (x, 2x + 3).
Step 2: Distance from origin to this point.
D = √[(x - 0)² + (2x + 3 - 0)²] = √[x² + (2x + 3)²]
Step 3: Minimize D² instead of D (same result, cleaner algebra).
D² = x² + (2x + 3)² = x² + 4x² + 12x + 9 = 5x² + 12x + 9
Step 4: Differentiate and set to zero.
D²' = 10x + 12 = 0
x = -1.2
Step 5: Find y.
y = 2(-1.2) + 3 = -2.4 + 3 = 0.6
Answer: The point (-1.2, 0.6)
Where Students Actually Lose Points
| Mistake | Why It Costs You | Fix |
|---|---|---|
| Forgetting to substitute the constraint | You end up with two variables and can't take a derivative | Always identify your constraint equation first |
| Not checking endpoints or boundary values | Critical points aren't always maxima or minima | Test the domain endpoints and any points where derivative is undefined |
| Solving for the wrong variable | You optimize cost when the problem asks for area | Circle what you're actually optimizing |
| Algebra errors in the substitution step | Everything after is wrong | Practice the substitution process until it's automatic |
| Forgetting the second derivative test or first derivative test | You can't confirm you found a max vs. min | Always verify your critical point gives you what the problem asks for |
The Second Derivative Test: When to Use It
Most optimization problems don't require it. You can usually tell from context whether you found a maximum or minimum:
- Area and volume problems: you're almost always looking for a maximum
- Cost problems: you're usually looking for a minimum
- Distance problems: you're looking for a minimum
That said, if the problem doesn't specify "maximum" or "minimum," use the second derivative test. It's faster than checking endpoints in most cases.
If f''(x) > 0 at your critical point, you have a minimum. If f''(x) < 0, you have a maximum.
What to Do the Night Before the Exam
Don't try to learn new content. Instead, practice setting up problems:
- Read a problem, write out the constraint equation and the function you need to optimize
- Don't solve it—just practice the setup
- This is the step most students struggle with, and it's what the exam tests
If you can consistently identify the constraint and write the primary equation, the calculus part takes care of itself.