Solving Optimization Problems — AP® Calculus AB

Part 1 · Hook

Why this matters

Packaging engineers squeeze the most volume out of a sheet of cardboard; utility planners route cable to spend the least money. Both run the same five-step routine: draw it, name the quantity, use the constraint, differentiate, and prove the answer is the best one.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. The objective function in an optimization problem is

the quantity to maximize or minimize, in one variable
the derivative set equal to zero
the domain of the problem
the second derivative

Show the answer

You optimize that function.

Correct: the quantity to maximize or minimize, in one variable:
the derivative set equal to zero:
the domain of the problem:
the second derivative:

2. $\frac{d}{dx}\sqrt{x^2 + 4} =$

$\frac{x}{\sqrt{x^2 + 4}}$
$\frac{1}{2\sqrt{x^2 + 4}}$
$2x\sqrt{x^2 + 4}$
$\frac{2x}{\sqrt{x^2 + 4}}$

Show the answer

Chain rule: $\frac{1}{2\sqrt{x^2 + 4}} \cdot 2x$ .

Correct: $\frac{x}{\sqrt{x^2 + 4}}$ :
$\frac{1}{2\sqrt{x^2 + 4}}$ :
$2x\sqrt{x^2 + 4}$ :
$\frac{2x}{\sqrt{x^2 + 4}}$ :

Part 4 · See it

See it first

Left: a 12-inch square sheet with a square of side x at each corner and dashed fold lines; folding gives an open box of height x and base 12 - 2x by 12 - 2x. Right: the graph of V(x) = x(12 - 2x) squared for 0 ≤ x ≤ 6, zero at both ends, highest at (2, 128). — Cutting squares of side $x$ from a 12-inch square sheet: $V(x) = x(12 - 2x)^2$ on $[0, 6]$ , largest at $x = 2$ , 128 cubic inches. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Draw the situation and label the variablesyou can see which quantities are fixed and which change
Write the objective function and the constraintthe constraint lets you eliminate a variable: the objective becomes a function of one variable
Find the domain from the physical limitsyou know the endpoints and which critical points count
Differentiate, solve, and compare candidates (or use one-critical-point reasoning)the absolute extremum, justified, stated in context with units

Part 6 · Key ideas

Key ideas

Constraint: a fixed condition (a perimeter, a volume, a budget) that links the variables. Use it to write the objective in one variable.
Steps: draw, label, objective, constraint, one variable, domain, critical points, justify, answer in context.
Reject critical points outside the domain (the cut $x = \frac{20}{3}$ inches on a 10-inch side).
Calculator problems: solve $C'(x) = 0$ with the solver or a graph, store the value, and compare candidates.

Part 7 · Misconception

A common mistake

The wrong idea: The objective function and the constraint are the same equation.

What actually happens: The constraint is the fixed fact (volume $= 32$ ); the objective is what you optimize (surface area). Using the constraint as if it were the objective, or forgetting to substitute it, is the most common setup error.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Diagram

An open box from a 16 by 10 sheet

A sheet of cardboard is 16 inches by 10 inches. A square of side

x

inches is cut from each corner, and the sides are folded up to make an open box.

A sheet 16 inches wide and 10 inches tall with a square of side x cut from each corner; the sides fold up along dashed lines into an open box of height x and base 16 - 2x by 10 - 2x inches. — LevlPrep original diagram.

1. Write the volume $V$ of the box, in cubic inches, as a function of $x$ .

Type an expression for $V(x) =$ .

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Height $x$ , base $(16 - 2x)$ by $(10 - 2x)$ : $V(x)$ $= x(16 - 2x)(10 - 2x)$ $= 4x^3 - 52x^2$ $+ 160x$ .

Answer: $V(x) =$ $x(16 - 2x) (10 - 2x)$

2. Which domain for $x$ makes sense?

$0 \le x \le 8$
$0 \le x \le 10$
$0 \le x \le 5$
$x \ge 0$

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Each dimension must be at least 0: $x \ge 0$ and $10 - 2x \ge 0$ , so $0 \le x \le 5$ . (The ends give zero volume.)

$0 \le x \le 8$ : At $x = 6$ , the side $10 - 2x$ is negative.
$0 \le x \le 10$ : Both sides would be negative long before $x = 10$ .
Correct: $0 \le x \le 5$ : The short side $10 - 2x$ must not be negative.
$x \ge 0$ : A cut larger than half the short side is impossible.

3. How large a square should be cut from each corner to make the box's volume as large as possible? Include units.

No calculator. Type the exact value (a fraction, π, e or a root is fine), then its unit.

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$V'(x)$ $= 12x^2 - 104x$ $+ 160$ $= 4(3x - 20)(x - 2)$ $= 0$ at $x = 2$ or $x = \frac{20}{3}$ . Only $x = 2$ inches is in $[0, 5]$ .

Answer: 2 inches

4. Find the largest possible volume of the box. Include units.

No calculator. Type the exact value (a fraction, π, e or a root is fine), then its unit.

Show the answer

$V(2) = 2 \cdot 12 \cdot 6 = 144$ cubic inches.

Answer: 144 cubic inches

5. Which justification shows that $x = 2$ gives the absolute maximum on $[0, 5]$ ?

$V'(2) = 0$ , so $V(2)$ is the largest volume
$x = 2$ is the smaller solution of $V'(x) = 0$
$V(2) = 144$ is a positive volume
$V$ is continuous on $[0, 5]$ ; $V(0) = 0$ , $V(2) = 144$ , $V(5) = 0$ , so the largest candidate is $V(2)$

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On the closed interval $[0, 5]$ , compare the candidates $V(0) = 0$ , $V(2) = 144$ and $V(5) = 0$ . The absolute maximum is 144 cubic inches at $x = 2$ .

$V'(2) = 0$ , so $V(2)$ is the largest volume: A critical point is only a candidate.
$x = 2$ is the smaller solution of $V'(x) = 0$ : Being the smaller root proves nothing about size.
$V(2) = 144$ is a positive volume: Positive does not mean largest.
Correct: $V$ is continuous on $[0, 5]$ ; $V(0) = 0$ , $V(2) = 144$ , $V(5) = 0$ , so the largest candidate is $V(2)$ : The Candidates Test: endpoints and the critical point in the domain.

6. An open box with a square base of side $x$ and height $h$ must hold 32 cubic feet. You want to use the least material. Which is the constraint?

$S = x^2 + 4xh$
$x^2 + 4xh = 32$
$4x + h = 32$
$x^2 h = 32$

Show the answer

The constraint is the fixed fact, the volume $x^2 h = 32$ ; it lets you write $h = \frac{32}{x^2}$ and the objective $S = x^2 + 4xh$ in one variable.

$S = x^2 + 4xh$ : That is the objective function, the amount of material to minimize.
$x^2 + 4xh = 32$ : 32 is a volume, not a surface area.
$4x + h = 32$ : That mixes lengths; 32 is a volume.
Correct: $x^2 h = 32$ : The fixed volume links $x$ and $h$ .

Part 9 · Summary

Summary

Solve an optimization problem in a fixed order: draw and label, write the objective function, use the constraint to reduce it to one variable, find the domain, find the critical points, justify the absolute extremum with candidates or one-critical-point reasoning, and answer in context with units.

Part 10 · Up next

What comes next

Part 11 · Connections