Chapter 10 — Quiz
10 questions on single-variable applied optimization. A mix of concept and computation. Try each before opening the answer; every answer cites the section to review. A scoring guide follows the last question.
1. In an optimization problem, what is the difference between the objective function and the constraint?
Answer
The **objective function** is the quantity you want to make largest or smallest (area, cost, profit, time) — it is what you differentiate. The **constraint** is a fixed relationship tying the variables together (a fixed perimeter, a fixed volume), and it is what you use to eliminate variables until the objective depends on one variable. A reliable tell: the constraint is the quantity that is *fixed* ("100 ft of fencing"); the objective is the quantity you are *free to push* ("maximize area"). (§10.2)2. State the closed-interval method for finding the global maximum and minimum of a continuous function on $[a, b]$.
Answer
(1) Find every critical point of $f$ in the open interval $(a, b)$ — where $f' = 0$ or $f'$ is undefined. (2) Evaluate $f$ at each critical point and at both endpoints $a$ and $b$. (3) The largest of these values is the global maximum; the smallest is the global minimum. The Extreme Value Theorem guarantees both exist, and no second-derivative test is needed — you simply compare a finite list. (§10.4)3. Find the global maximum and minimum of $f(x) = 2x^3 - 9x^2 + 12x$ on $[0, 3]$.
Answer
$f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x-1)(x-2) = 0 \implies x = 1, 2$. Candidates: $f(0) = 0$, $f(1) = 2 - 9 + 12 = 5$, $f(2) = 16 - 36 + 24 = 4$, $f(3) = 54 - 81 + 36 = 9$. Global **max** $= 9$ at $x = 3$ (an endpoint!); global **min** $= 0$ at $x = 0$. The interior local max at $x=1$ loses to the endpoint. (§10.4)4. A farmer has $80$ ft of fencing and builds a rectangular pen against a wall (three sides fenced). What dimensions maximize the area, and what is that area?
Answer
Perpendicular sides $x$, parallel side $y$: $2x + y = 80 \implies y = 80 - 2x$. $A(x) = x(80 - 2x) = 80x - 2x^2$, domain $[0, 40]$. $A'(x) = 80 - 4x = 0 \implies x = 20$, $y = 40$. $A(20) = 20 \cdot 40 = 800$ ft². The parallel side ($40$) is twice each perpendicular side ($20$) — the §10.5 signature. Max area $\boxed{800 \text{ ft}^2}$. (§10.5)5. When finding the point on a curve closest to a given point, why do we minimize the squared distance rather than the distance itself?
Answer
Because $\sqrt{\,\cdot\,}$ is an increasing function, $D$ and $D^2$ attain their minimum at the *same* input. Minimizing $D^2$ avoids differentiating a square root (which produces messy chain-rule fractions); the squared distance is a polynomial with an easy derivative. This "minimize the square" move works for any distance, time, or magnitude objective and should be reflexive. (§10.7)6. A closed cylindrical can of fixed volume is designed to minimize surface area. What is the relationship between its optimal height and radius, and why don't real cans follow it?
Answer
The minimum-material can has **height equal to diameter**, $h = 2r$. Real cans are taller because the pure-material objective ignores the thicker, costlier top lid, the seams, shipping a squat shape, and how a hand grips it. When those real costs enter the objective function, the optimum shifts toward a taller cylinder — the calculus is right, but the objective must capture every cost that matters. (§10.6)7. State the economic rule that characterizes the profit-maximizing output, and explain why it follows from setting $P'(x) = 0$.
Answer
**Marginal revenue equals marginal cost:** $R'(x) = C'(x)$. Since profit is $P = R - C$, the condition $P'(x) = 0$ is exactly $R'(x) - C'(x) = 0$. Produce up to the point where the revenue from the next unit equals its cost; beyond it each extra unit loses money, before it you leave money on the table. Confirm with $P'' < 0$ for a maximum. (§10.10)8. A retailer faces annual demand $D$, per-order cost $K$, and holding cost $h$ per unit per year. Write the total-cost function $T(Q)$ and the economic order quantity $Q^*$.
Answer
$T(Q) = \dfrac{DK}{Q} + \dfrac{hQ}{2}$ for $Q > 0$: the first term is ordering cost ($D/Q$ orders at $K$ each), the second is holding cost (average inventory $Q/2$ at $h$ each). Setting $T'(Q) = -\dfrac{DK}{Q^2} + \dfrac{h}{2} = 0$ gives $Q^* = \sqrt{\dfrac{2DK}{h}}$. Since $T'' = 2DK/Q^3 > 0$, this is the global minimum. (§10.11)9. Find the critical point of $f(x) = x^2 + \dfrac{16}{x}$ for $x > 0$, classify it, and explain how you know it is a global minimum even though the domain is not a closed interval.
Answer
$f'(x) = 2x - \dfrac{16}{x^2} = 0 \implies 2x^3 = 16 \implies x^3 = 8 \implies x = 2$. $f''(x) = 2 + \dfrac{32}{x^3} > 0$ for all $x > 0$, so $f$ is convex (concave up) everywhere and the critical point is a minimum. Because $f \to +\infty$ as $x \to 0^+$ (the $16/x$ term) and as $x \to \infty$ (the $x^2$ term), the lone interior critical point must be the **global** minimum — squeezed between two infinities. $f(2) = 4 + 8 = 12$. (§10.6, §10.15)10. Among all rectangles with a fixed perimeter $P$, the square has the shortest diagonal. A student concludes the square therefore has the longest diagonal too, since it is "the optimal rectangle." What is the error, and where does the longest diagonal actually occur?
Answer
The error is assuming a critical point answers every extremal question. Minimizing $d^2 = a^2 + (\tfrac P2 - a)^2$ gives $a = P/4$ (a square) with $f'' > 0$ — a **minimum** of the diagonal, not a maximum. The *longest* diagonal occurs nowhere in the interior: it is approached at the degenerate boundary, as the rectangle collapses to a sliver ($a \to 0$ or $a \to P/2$) and the diagonal stretches toward $P/2$. A critical point is a candidate, never an answer — confirm its type and check the boundary. (§10.13, §10.15)Scoring Guide
| Score | Interpretation |
|---|---|
| 9–10 | Excellent. You can set up, solve, and justify optimization problems, and you respect the difference between a candidate and an answer. Ready for Chapter 11. |
| 7–8 | Solid. Review whichever of the closed-interval method (§10.4), the global-extremum tests (§10.15), or the economic rules (§10.10–§10.11) tripped you up. |
| 5–6 | Partial. The mechanics are there but the setup (objective vs. constraint, §10.2–§10.3) or the justification (max vs. min, §10.13) needs another pass. Rework the §10.5–§10.6 examples by hand. |
| 0–4 | Reread the chapter, focusing on §10.2–§10.4. Redo the worked pen and can problems from scratch before retrying — the translation skill is the whole game. |
A perfect score is not the goal; an explicit global-extremum justification on every problem is. That habit, more than any single answer, is what Chapter 10 is teaching.