Solving Optimization Problems — AP® Calculus AB Notes

The introduction showed the idea; this page is the full routine, applied to the kinds of problems the exam uses: boxes, cans, cables, shapes inside shapes. The calculus is short. Most mistakes happen in the setup, so the setup gets most of the attention here.

The routine

Draw the situation and label every changing length or amount with a letter.
Write the objective function: the quantity to maximize or minimize.
Write the constraint: the fixed fact that links the variables (a volume, a perimeter, a total length).
Use the constraint to write the objective in one variable.
Find the domain from the physical limits.
Find the critical points in the domain.
Justify the absolute extremum: candidates on a closed domain, or one critical point that is a relative extremum.
Answer the question in context, with units.

Keep the objective and the constraint apart. In "an open box with a square base must hold 32 cubic feet; use the least material," the constraint is the volume $x^2 h = 32$ , and the objective is the material, $S = x^2 + 4xh$ . Mixing them up is the most common setup error.

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). — **Figure 1.** Corner squares of side $x$ cut from a 12-inch sheet: the volume $V(x) = x(12 - 2x)^2$ is zero at both ends of $[0, 6]$ and largest at $x = 2$ . LevlPrep original diagram.

Worked example 1. A 16-inch by 10-inch sheet has a square of side $x$ cut from each corner, and the sides are folded up into an open box. Find the largest volume.

Objective: the base loses $2x$ in each direction, so $V(x)$ $= x(16 - 2x)(10 - 2x)$ $= 4x^3 - 52x^2$ $+ 160x$ .

Domain: the short side needs $10 - 2x \ge 0$ , so $0 \le x \le 5$ .

Critical points: $V'(x)$ $= 12x^2 - 104x$ $+ 160$ $= 4(3x - 20)(x - 2)$ , zero at $x = 2$ and $x = \frac{20}{3}$ . Only $x = 2$ is in the domain; $\frac{20}{3}$ would cut away more than the sheet has.

Justify: $V(0) = 0$ , $V(2) = 2 \cdot 12 \cdot 6 = 144$ , $V(5) = 0$ .

Answer: cut 2-inch squares; the box holds 144 cubic inches.

Worked example 2. A closed can must hold 500 cubic centimeters. What radius uses the least metal?

Objective: surface area, two ends plus the side: $S = 2\pi r^2 + 2\pi r h$ .

Constraint: $\pi r^2 h = 500$ , so $h = \frac{500}{\pi r^2}$ and $S(r) = 2\pi r^2 + \frac{1000}{r}$ , for $r > 0$ .

Critical point: $S'(r)$ $= 4\pi r - \frac{1000}{r^2}$ $= 0$ gives $r^3 = \frac{250}{\pi}$ , $r = \sqrt[3]{\frac{250}{\pi}} \approx 4.301$ cm.

Justify: it is the only critical point for $r > 0$ , and $S''(r) = 4\pi + \frac{2000}{r^3} > 0$ , so it is a relative minimum and therefore the absolute minimum.

The domain $r > 0$ has no endpoints to compare, which is exactly when the one-critical-point argument is needed.

Calculator optimization

When the critical point has no clean formula, use the calculator to solve $f'(x) = 0$ , store the result, and evaluate.

Worked example 3. A cable runs from a station on one bank of a river 2 km wide, underwater to a point $x$ km downstream on the far bank, then along the bank to a factory 8 km downstream. It costs 5 thousand dollars per km underwater and 3.5 on land, so $C(x)$ $= 5\sqrt{x^2 + 4} + 3.5(8 - x)$ , $0 \le x \le 8$ .

$C'(x)$ $= \frac{5x}{\sqrt{x^2 + 4}} - 3.5$ (chain rule on the square root). Solving $C'(x) = 0$ gives $x \approx 1.960$ .

Candidates: $C(0) = 38$ , $C(1.960) \approx 35.141$ , $C(8) = 5\sqrt{68} \approx 41.231$ . The cheapest route lands about 1.960 km downstream and costs about 35.141 thousand dollars.

Shapes inside shapes

For a rectangle with its base on the diameter of a semicircle of radius 5, put the top right corner at $(x, \sqrt{25 - x^2})$ . The width is $2x$ , so $A(x) = 2x\sqrt{25 - x^2}$ on $[0, 5]$ . The derivative is zero at $x = \frac{5}{\sqrt{2}}$ , and $A\left(\frac{5}{\sqrt{2}}\right) = 25$ , larger than $A(0) = A(5) = 0$ . Placing a corner on the curve is what links the variables here; that is the constraint.

Checklist before you write the answer

Is each critical point inside the domain? Reject the others, and say why.
Did you justify the extremum, not just find $f'(x) = 0$ ?
Did you answer what was asked (the dimensions, the cost, the radius) with units?
On a calculator problem, did you show the setup, $C'(x) = 0$ , and give three decimal places?