Chapter 10 — Exercises

42 problems on single-variable applied optimization. Every problem is set up and solved: identify the objective and the constraint, use the constraint to reduce to one variable, find and classify the critical point, and confirm it is a global extremum (closed-interval method, boundary analysis, or a concavity/sign argument). Difficulty runs ⭐ (one-step) to ⭐⭐⭐⭐ (multi-stage modeling). The setup is the hard part; the calculus is mechanical (§10.1). Work each problem before opening the solution.

How to Use These Problems

The problems are grouped by the kind of application, mirroring the chapter's tour through geometry, materials, distance and time, economics, and biology. Within each part the difficulty climbs. A four-star problem is not harder calculus — it is harder translation: more variables to eliminate, a trickier domain, or a result you must interpret. Follow the eight-step procedure of §10.3 every time, and always make the global-extremum justification an explicit step (§10.15).

Difficulty Distribution

Tier Meaning Count
⭐ One-step: objective already in one variable, or a direct closed-interval check 9
⭐⭐ Standard: set up objective + constraint, eliminate, classify 16
⭐⭐⭐ Multi-step: messier algebra, careful domain, or interpretation 13
⭐⭐⭐⭐ Extended modeling: derive a general formula or combine several ideas 4
Total 42

Part A — Critical Points and the Closed-Interval Method (§10.4)

A1. ⭐ Find the global maximum and minimum of $f(x) = x^2 - 6x + 5$ on $[0, 5]$.

Solution $f'(x) = 2x - 6 = 0 \implies x = 3$, inside $[0,5]$. Candidates: $f(0) = 5$, $f(3) = 9 - 18 + 5 = -4$, $f(5) = 25 - 30 + 5 = 0$. Global **max** $= 5$ at $x = 0$; global **min** $= -4$ at $x = 3$. (§10.4)

A2. ⭐ Find the global extrema of $f(x) = x^3 - 3x^2$ on $[-1, 3]$.

Solution $f'(x) = 3x^2 - 6x = 3x(x-2) = 0 \implies x = 0, 2$, both inside. Candidates: $f(-1) = -1 - 3 = -4$, $f(0) = 0$, $f(2) = 8 - 12 = -4$, $f(3) = 27 - 27 = 0$. Global **max** $= 0$ (at $x = 0$ and $x = 3$); global **min** $= -4$ (at $x = -1$ and $x = 2$). The max and min each *tie* between an interior point and an endpoint — exactly why all four candidates must be checked. (§10.4)

A3. ⭐ Find the global extrema of $f(x) = x^4 - 8x^2$ on $[-1, 3]$.

Solution $f'(x) = 4x^3 - 16x = 4x(x^2 - 4) = 4x(x-2)(x+2) = 0 \implies x = 0, 2$ (note $x=-2 \notin [-1,3]$). Candidates: $f(-1) = 1 - 8 = -7$, $f(0) = 0$, $f(2) = 16 - 32 = -16$, $f(3) = 81 - 72 = 9$. Global **max** $= 9$ at $x = 3$; global **min** $= -16$ at $x = 2$. (§10.4)

A4. ⭐⭐ Find the global extrema of $f(x) = x + \dfrac{4}{x}$ on $[1, 4]$.

Solution $f'(x) = 1 - \dfrac{4}{x^2} = 0 \implies x^2 = 4 \implies x = 2$ (only $x=2$ is in $[1,4]$). Candidates: $f(1) = 1 + 4 = 5$, $f(2) = 2 + 2 = 4$, $f(4) = 4 + 1 = 5$. Global **min** $= 4$ at $x = 2$; global **max** $= 5$ (at $x = 1$ and $x = 4$). (§10.4)

A5. ⭐⭐ A continuous function on $[0, 10]$ has exactly one critical point, at $x = 4$, where the second-derivative test shows a local maximum. Is $x=4$ necessarily the global maximum? Explain.

Solution **Yes for the maximum, here.** With a single interior critical point that is a local max, a continuous function rises to $x=4$ then falls, so there is nowhere higher and the global *maximum* is $f(4)$ (the "single critical point" rule, §10.15). But the global *minimum* still sits at an endpoint — the smaller of $f(0)$ and $f(10)$ — so you must check those. Beware the mirror case: a single interior local *minimum* forces the global maximum to an endpoint. Always state which extremum you have. (§10.4, §10.15)

Part B — Maximum Area and Volume (§10.5, §10.9)

B1. ⭐ A rectangle has perimeter $40$ m. What dimensions maximize its area?

Solution Constraint $2x + 2y = 40 \implies y = 20 - x$. Objective $A = xy = x(20 - x) = 20x - x^2$, domain $[0, 20]$. $A'(x) = 20 - 2x = 0 \implies x = 10$, so $y = 10$. $A''= -2 < 0$, a max. The optimum is a **square** $10 \times 10$ with area $100$ m². (§10.5)

B2. ⭐⭐ A farmer has $200$ ft of fencing for a rectangular pen against a barn wall (no fence on the wall side). Maximize the enclosed area.

Solution Let $x$ be each side perpendicular to the wall and $y$ the parallel side. Constraint $2x + y = 200 \implies y = 200 - 2x$. Objective $A(x) = x(200 - 2x) = 200x - 2x^2$, domain $[0, 100]$. $A'(x) = 200 - 4x = 0 \implies x = 50$, $y = 100$. Candidates: $A(0) = 0$, $A(100) = 0$, $A(50) = 50 \cdot 100 = 5000$ ft². Optimum $50 \times 100$, area $\boxed{5000 \text{ ft}^2}$. As in §10.5, the parallel side is twice each perpendicular side. (§10.5)

B3. ⭐⭐ A rectangular field is divided into two equal pens by an internal fence parallel to one pair of sides. With $300$ m of fencing total, maximize the total area.

Solution Let $y$ be the side that the dividing fence is parallel to (so three segments of length $y$: two outer plus one divider) and $x$ the other side (two segments). Constraint $2x + 3y = 300 \implies x = 150 - \tfrac32 y$. Objective $A = xy = (150 - \tfrac32 y)y = 150y - \tfrac32 y^2$, domain $[0,100]$. $A'(y) = 150 - 3y = 0 \implies y = 50$, $x = 75$. $A(50) = 75\cdot 50 = 3750$ m². Optimum area $\boxed{3750 \text{ m}^2}$. (§10.5)

B4. ⭐ From an $18$-inch square sheet, cut squares of side $x$ from the corners and fold up to make an open box. Write $V(x)$ and state its domain.

Solution Base side $= 18 - 2x$, height $= x$, so $V(x) = x(18 - 2x)^2$ on $[0, 9]$. (Solving is B5.) (§10.9)

B5. ⭐⭐ Maximize the volume of the open box in B4.

Solution $V(x) = x(18 - 2x)^2 = x(324 - 72x + 4x^2) = 4x^3 - 72x^2 + 324x$. $V'(x) = 12x^2 - 144x + 324 = 12(x^2 - 12x + 27) = 12(x-3)(x-9) = 0 \implies x = 3$ or $x = 9$. Candidates on $[0,9]$: $V(0) = 0$, $V(9) = 0$, $V(3) = 3(18-6)^2 = 3 \cdot 144 = 432$ in³. Optimum $\boxed{432 \text{ in}^3}$ with $3$-inch cuts, a $12 \times 12 \times 3$ box. The endpoint $x=9$ is a critical point that is *not* the answer (§10.9). (§10.9)

B6. ⭐⭐⭐ A rectangle is inscribed with its base on the $x$-axis and its top two corners on the parabola $y = 12 - x^2$ ($y \ge 0$). Maximize its area.

Solution By symmetry the corners are at $(\pm x, 12 - x^2)$ with $x > 0$. Width $= 2x$, height $= 12 - x^2$, so $A(x) = 2x(12 - x^2) = 24x - 2x^3$, domain $[0, \sqrt{12}]$. $A'(x) = 24 - 6x^2 = 0 \implies x^2 = 4 \implies x = 2$. Then height $= 12 - 4 = 8$, width $= 4$, area $= 32$. Endpoints give $A = 0$, so the optimum is $\boxed{32}$. (§10.5, §10.9)

B7. ⭐⭐⭐ A Norman window is a rectangle topped by a semicircle. The total perimeter is $P = 12$ m. Find the radius $r$ of the semicircle that maximizes the window's area (lets in the most light).

Solution Let the rectangle be width $2r$ (matching the semicircle's diameter) and height $y$. Perimeter $= 2r + 2y + \pi r = 12$ (bottom, two sides, semicircular arc), so $y = \dfrac{12 - 2r - \pi r}{2}$. Area $= 2r y + \tfrac12 \pi r^2 = 2r \cdot \dfrac{12 - 2r - \pi r}{2} + \tfrac12 \pi r^2 = 12r - 2r^2 - \pi r^2 + \tfrac12\pi r^2 = 12r - 2r^2 - \tfrac12\pi r^2$. $A'(r) = 12 - 4r - \pi r = 0 \implies r = \dfrac{12}{4 + \pi} \approx 1.68$ m. $A'' = -4 - \pi < 0$, a max. So the light-maximizing radius is $\boxed{r = 12/(4+\pi)}$. (§10.5)

B8. ⭐⭐⭐⭐ A cylinder is inscribed in a sphere of fixed radius $R$. Find the height $h$ of the cylinder of maximum volume, and show the ratio $h/R$ is independent of $R$.

Solution Let the cylinder have radius $\rho$ and height $h$. Inscribed in the sphere, a right triangle from the center gives $\rho^2 + (h/2)^2 = R^2$, so $\rho^2 = R^2 - h^2/4$. Volume $V = \pi \rho^2 h = \pi\left(R^2 - \tfrac{h^2}{4}\right)h = \pi R^2 h - \tfrac{\pi}{4}h^3$, domain $[0, 2R]$. $V'(h) = \pi R^2 - \tfrac{3\pi}{4}h^2 = 0 \implies h^2 = \tfrac{4R^2}{3} \implies h = \dfrac{2R}{\sqrt 3}$. $V'' = -\tfrac{3\pi}{2}h < 0$, a max. Thus $h/R = 2/\sqrt 3 \approx 1.155$, **independent of $R$** — a scale-free optimum, the kind of clean structural fact §10.5 highlights. (§10.9, §10.15)

Part C — Minimum Material and Cost (§10.6)

C1. ⭐ A closed cylindrical can must hold $V_0 = 500$ cm³. Write the surface-area objective $A(r)$ after eliminating the height.

Solution Constraint $\pi r^2 h = 500 \implies h = 500/(\pi r^2)$. $A = 2\pi r^2 + 2\pi r h = 2\pi r^2 + \dfrac{1000}{r}$, $r > 0$. (Solving is C2.) (§10.6)

C2. ⭐⭐ Minimize the surface area of the can in C1.

Solution $A'(r) = 4\pi r - \dfrac{1000}{r^2} = 0 \implies 4\pi r^3 = 1000 \implies r^3 = \dfrac{250}{\pi} \implies r = (250/\pi)^{1/3} \approx 4.30$ cm. $A''(r) = 4\pi + 2000/r^3 > 0$, convex, so a global min (boundary analysis: $A \to \infty$ as $r \to 0^+$ and $r \to \infty$). Using $\pi r^3 = 250$ gives $\pi r^2 = 250/r$, so $h = 500/(\pi r^2) = 500/(250/r) = 2r$: **height equals diameter**, just as in §10.6. (§10.6, §10.15)

C3. ⭐⭐ An open-topped box has a square base of side $x$ and must hold $32$ ft³. Minimize the material (base plus four sides).

Solution Volume $x^2 h = 32 \implies h = 32/x^2$. Material $S = x^2 + 4xh = x^2 + 4x\cdot\dfrac{32}{x^2} = x^2 + \dfrac{128}{x}$, $x > 0$. $S'(x) = 2x - \dfrac{128}{x^2} = 0 \implies x^3 = 64 \implies x = 4$ ft, $h = 32/16 = 2$ ft. $S'' = 2 + 256/x^3 > 0$, a min. Material $= 16 + 32 = 48$ ft². (§10.6)

C4. ⭐⭐⭐ A closed rectangular box has a square base and volume $V_0$. The top and bottom material costs $\$2$/ft² and the four sides cost $\$1$/ft². Find the base side $x$ minimizing total cost.

Solution Base side $x$, height $h$, $x^2 h = V_0 \implies h = V_0/x^2$. Cost $= 2(2x^2) + 1(4xh) = 4x^2 + 4x\cdot\dfrac{V_0}{x^2} = 4x^2 + \dfrac{4V_0}{x}$. $C'(x) = 8x - \dfrac{4V_0}{x^2} = 0 \implies x^3 = \dfrac{V_0}{2} \implies x = (V_0/2)^{1/3}$. $C'' = 8 + 8V_0/x^3 > 0$, a min. The pricier top/bottom pushes the optimum toward a *smaller-footprint, taller* box than the equal-cost case. (§10.6)

C5. ⭐⭐⭐ A pipeline runs from a refinery on one bank of a straight $400$-m-wide river to a tank $1000$ m downstream on the opposite bank. Pipe under the river costs $\$500$/m; pipe along the bank costs $\$300$/m. The line goes straight under the river to a point $x$ m downstream on the far bank, then along the bank. Minimize cost.

Solution Underwater length $= \sqrt{400^2 + x^2}$; along-bank length $= 1000 - x$. Cost $C(x) = 500\sqrt{160000 + x^2} + 300(1000 - x)$, $x \in [0, 1000]$. $C'(x) = \dfrac{500x}{\sqrt{160000 + x^2}} - 300 = 0 \implies 500x = 300\sqrt{160000 + x^2} \implies 25x^2 = 9(160000 + x^2) \implies 16x^2 = 1440000 \implies x^2 = 90000 \implies x = 300$ m. Check candidates: $C(300) = 500\sqrt{160000+90000} + 300(700) = 500\cdot 500 + 210000 = \$460{,}000$. $C(0) = 500\cdot 400 + 300\cdot 1000 = \$500{,}000$; $C(1000) = 500\sqrt{1{,}160{,}000} \approx \$538{,}500$. Optimum $\boxed{x = 300 \text{ m},\ \$460{,}000}$. (§10.6, §10.8)

C6. ⭐⭐ A poster must contain $384$ cm² of printed area, with margins of $4$ cm on top and bottom and $2$ cm on each side. Find the overall dimensions that minimize the total area of the poster.

Solution Let the printed region be $x$ wide and $y$ tall: $xy = 384$. Overall poster is $(x + 4)$ wide and $(y + 8)$ tall. Total area $S = (x+4)(y+8)$ with $y = 384/x$: $S = (x+4)\left(\dfrac{384}{x} + 8\right) = 384 + 8x + \dfrac{1536}{x} + 32 = 416 + 8x + \dfrac{1536}{x}$. $S'(x) = 8 - \dfrac{1536}{x^2} = 0 \implies x^2 = 192 \implies x = \sqrt{192} = 8\sqrt3 \approx 13.86$ cm, $y = 384/x \approx 27.7$ cm. Overall $\approx 17.9 \times 35.7$ cm. $S'' = 3072/x^3 > 0$, a min. (§10.6)

Part D — Distance and Time (§10.7, §10.8)

D1. ⭐⭐ Find the point on the line $y = 2x + 3$ closest to the origin. (Minimize squared distance.)

Solution Point $(x, 2x+3)$; $D^2 = x^2 + (2x+3)^2 = x^2 + 4x^2 + 12x + 9 = 5x^2 + 12x + 9$. $\dfrac{d(D^2)}{dx} = 10x + 12 = 0 \implies x = -\tfrac{6}{5}$, $y = 2(-\tfrac65) + 3 = \tfrac35$. Closest point $\left(-\tfrac65, \tfrac35\right)$. ($D^2{}'' = 10 > 0$, a min.) Minimizing $D^2$ rather than $D$ keeps the algebra polynomial (§10.7). (§10.7)

D2. ⭐⭐⭐ Find the point on the parabola $y = x^2$ closest to $(6, 3)$.

Solution Minimize $f(x) = (x - 6)^2 + (x^2 - 3)^2 = x^2 - 12x + 36 + x^4 - 6x^2 + 9 = x^4 - 5x^2 - 12x + 45$. $f'(x) = 4x^3 - 10x - 12 = 2(2x^3 - 5x - 6)$. The cubic $2x^3 - 5x - 6$ has no rational root (testing $x = 2$ gives $16 - 10 - 6 = 0$ — wait, recheck: $2(8) - 5(2) - 6 = 16 - 10 - 6 = 0$). So $x = 2$ **is** a root! Factor: $2x^3 - 5x - 6 = (x - 2)(2x^2 + 4x + 3)$, and the quadratic $2x^2 + 4x + 3$ has discriminant $16 - 24 < 0$, no real roots. So the only critical point is $x = 2$, $y = 4$. Since $f \to \infty$ as $x \to \pm\infty$ and this is the sole critical point, it is the global min. Closest point $\boxed{(2, 4)}$, at distance $\sqrt{(2-6)^2 + (4-3)^2} = \sqrt{17}$. (§10.7)

D3. ⭐⭐⭐ Find the point on the curve $y = \sqrt{x}$ closest to the point $(4, 0)$.

Solution Point $(x, \sqrt x)$, $x \ge 0$. $f(x) = (x - 4)^2 + (\sqrt x)^2 = (x-4)^2 + x = x^2 - 8x + 16 + x = x^2 - 7x + 16$. $f'(x) = 2x - 7 = 0 \implies x = \tfrac72$, $y = \sqrt{7/2} \approx 1.87$. $f'' = 2 > 0$, a min. Closest point $\left(\tfrac72, \sqrt{7/2}\right)$. (§10.7)

D4. ⭐⭐⭐ A lifeguard is at point $A$ on the beach, $40$ m back from a straight shoreline. A swimmer is $30$ m out, with horizontal separation $L = 100$ m from $A$ along the shore. The guard runs at $5$ m/s on sand and swims at $2$ m/s. Set up $T(x)$ and write the optimality condition (Snell's-law form). You need not solve numerically.

Solution Let $x$ be the along-shore distance from the foot of $A$ to the entry point. Run $\sqrt{40^2 + x^2}$ on sand, swim $\sqrt{30^2 + (100 - x)^2}$. $T(x) = \dfrac{\sqrt{1600 + x^2}}{5} + \dfrac{\sqrt{900 + (100-x)^2}}{2}$, $x \in [0, 100]$. $T'(x) = \dfrac{x}{5\sqrt{1600 + x^2}} - \dfrac{100 - x}{2\sqrt{900 + (100-x)^2}} = 0$, i.e. $\dfrac{\sin\theta_1}{5} = \dfrac{\sin\theta_2}{2}$ — **Snell's law** (§10.8). Since the guard is faster on sand, $\sin\theta_1 > \sin\theta_2$, so $\theta_1 > \theta_2$: the path bends toward the perpendicular once in the slow water. (§10.8)

D5. ⭐ A projectile launched at angle $\theta$ with fixed speed $v$ on level ground has range $R(\theta) = \dfrac{v^2 \sin(2\theta)}{g}$. Find the angle maximizing range, $0 \le \theta \le \pi/2$.

Solution $R'(\theta) = \dfrac{2v^2}{g}\cos(2\theta) = 0 \implies \cos(2\theta) = 0 \implies 2\theta = \tfrac{\pi}{2} \implies \theta = \tfrac{\pi}{4} = 45°$. $R'' = -\tfrac{4v^2}{g}\sin(2\theta) < 0$ at $2\theta = \pi/2$, a max. The classic $45°$ result. (§10.8)

Part E — Economic Optimization (§10.10, §10.11)

E1. ⭐⭐ A firm has revenue $R(x) = 200x - 2x^2$ and cost $C(x) = 40x + 300$. Find the profit-maximizing output and the maximum profit.

Solution $P(x) = R - C = 200x - 2x^2 - 40x - 300 = -2x^2 + 160x - 300$. $P'(x) = -4x + 160 = 0 \implies x = 40$. $P'' = -4 < 0$, a max. $P(40) = -2(1600) + 6400 - 300 = -3200 + 6400 - 300 = 2900$. Check MR = MC: $R'(x) = 200 - 4x$, $R'(40) = 40$; $C'(x) = 40$. ✓ Max profit $\boxed{\$2900}$ at $x = 40$. (§10.10)

E2. ⭐ A monopolist faces demand $p = 50 - x$ (price per unit). Find the output maximizing total revenue $R = px$.

Solution $R(x) = (50 - x)x = 50x - x^2$. $R'(x) = 50 - 2x = 0 \implies x = 25$, $p = 25$, $R = 625$. $R'' = -2 < 0$, a max. Revenue peaks where marginal revenue $R' = 0$. (§10.10)

E3. ⭐⭐ With demand $p = 100 - 2x$ and cost $C(x) = 20x + 50$, find the profit-maximizing output.

Solution $R = px = 100x - 2x^2$; $P = R - C = 100x - 2x^2 - 20x - 50 = -2x^2 + 80x - 50$. $P'(x) = -4x + 80 = 0 \implies x = 20$. $p = 100 - 40 = 60$, $P(20) = -800 + 1600 - 50 = 750$. MR $= 100 - 4x = 20 =$ MC at $x=20$. ✓ (§10.10)

E4. ⭐⭐ A retailer sells $D = 4000$ units/year, pays $K = \$25$ per order, and holds inventory at $h = \$2$/unit/year. Find the economic order quantity $Q^*$ and the minimum total annual cost.

Solution $Q^* = \sqrt{\dfrac{2DK}{h}} = \sqrt{\dfrac{2 \cdot 4000 \cdot 25}{2}} = \sqrt{100000} \approx 316.2$ units. At the EOQ the two costs are equal, each $= \dfrac{hQ^*}{2} = \dfrac{2 \cdot 316.2}{2} = 316.2$, so total cost $T \approx \$632.5$/year. (§10.11)

E5. ⭐⭐⭐⭐ Derive the EOQ formula from scratch, prove the optimum is a global minimum, and show that at $Q^*$ the annual ordering cost equals the annual holding cost.

Solution Annual cost $T(Q) = \dfrac{DK}{Q} + \dfrac{hQ}{2}$, $Q > 0$ (§10.11). $T'(Q) = -\dfrac{DK}{Q^2} + \dfrac{h}{2} = 0 \implies Q^2 = \dfrac{2DK}{h} \implies Q^* = \sqrt{\dfrac{2DK}{h}}$. $T''(Q) = \dfrac{2DK}{Q^3} > 0$ for all $Q > 0$, so $T$ is convex and $Q^*$ is the **global** minimum (also $T \to \infty$ at both boundaries). At $Q^*$: ordering cost $= \dfrac{DK}{Q^*} = \dfrac{DK}{\sqrt{2DK/h}} = \sqrt{\dfrac{DKh}{2}}$; holding cost $= \dfrac{hQ^*}{2} = \dfrac{h}{2}\sqrt{\dfrac{2DK}{h}} = \sqrt{\dfrac{DKh}{2}}$. **Equal** — the optimum splits cost evenly between the two competing forces, the hallmark of a balanced trade-off. (§10.11, §10.15)

E6. ⭐⭐⭐ A firm's average cost per unit is $\bar{C}(x) = \dfrac{C(x)}{x}$ where $C(x) = x^2 + 16x + 400$. Find the output minimizing average cost, and show that there marginal cost equals average cost.

Solution $\bar C(x) = \dfrac{x^2 + 16x + 400}{x} = x + 16 + \dfrac{400}{x}$. $\bar C'(x) = 1 - \dfrac{400}{x^2} = 0 \implies x^2 = 400 \implies x = 20$. $\bar C'' = 800/x^3 > 0$, a min. $\bar C(20) = 20 + 16 + 20 = 56$. Marginal cost $C'(x) = 2x + 16$, $C'(20) = 56 = \bar C(20)$. ✓ Average cost is minimized where the marginal-cost curve crosses it — a standard microeconomic fact, and a tidy consequence of $\frac{d}{dx}\frac{C}{x}=0 \iff C' = C/x$. (§10.10)

E7. ⭐⭐⭐ A tour operator charges $\$800$ per person if exactly $100$ people sign up. For each additional person above $100$, the price per person drops by $\$4$ (group discount), up to a bus capacity of $180$. How many passengers maximize total revenue?

Solution Let $x$ be passengers above $100$, $0 \le x \le 80$. Headcount $= 100 + x$; price $= 800 - 4x$. Revenue $R(x) = (100 + x)(800 - 4x) = 80000 - 400x + 800x - 4x^2 = 80000 + 400x - 4x^2$. $R'(x) = 400 - 8x = 0 \implies x = 50$, i.e. $150$ passengers at $\$600$ each. $R(50) = 150 \cdot 600 = \$90{,}000$. Endpoints: $R(80) = 180 \cdot 480 = \$86{,}400$, $R(0)=\$80{,}000$. Optimum $\boxed{150 \text{ passengers}}$. (§10.10)

Part F — Biological Optimization (§10.12)

F1. ⭐⭐ A forager's cumulative energy gain in a patch is $g(t) = \dfrac{20t}{t + 3}$ after time $t$. The average travel time between patches is $\tau = 2$. Find the optimal residence time $t^*$ via the marginal value theorem $g'(t^*) = \dfrac{g(t^*)}{\tau + t^*}$.

Solution $g'(t) = \dfrac{20(t+3) - 20t}{(t+3)^2} = \dfrac{60}{(t+3)^2}$. The condition $g'(t) = \dfrac{g(t)}{\tau + t}$ becomes $$\dfrac{60}{(t+3)^2} = \dfrac{20t/(t+3)}{2 + t} = \dfrac{20t}{(t+3)(2+t)}.$$ Multiply both sides by $(t+3)^2(2+t)$: $60(2 + t) = 20t(t+3) \implies 120 + 60t = 20t^2 + 60t \implies 20t^2 = 120 \implies t^2 = 6 \implies t^* = \sqrt 6 \approx 2.45$. The bird should leave after about $2.45$ time units. (§10.12)

F2. ⭐⭐⭐ For the same gain curve $g(t) = \dfrac{20t}{t+3}$, suppose travel time rises to $\tau = 8$ (patches farther apart). Show the optimal residence time increases, and explain why.

Solution Now $\dfrac{60}{(t+3)^2} = \dfrac{20t}{(t+3)(8+t)} \implies 60(8 + t) = 20t(t+3) \implies 480 + 60t = 20t^2 + 60t \implies 20t^2 = 480 \implies t^2 = 24 \implies t^* = \sqrt{24} \approx 4.90$. The residence time roughly **doubles** versus F1. **Why:** when travel is costly, abandoning a patch is expensive, so the optimal strategy is to stay longer and squeeze more from each patch before paying the travel toll again — exactly the marginal value theorem's prediction (§10.12). (§10.12)

F3. ⭐⭐⭐⭐ Optimal clutch size. A bird laying $n$ eggs has each chick survive with probability $s(n) = 1 - \dfrac{n}{20}$ (more chicks means thinner feeding). The expected number of surviving offspring is $W(n) = n \cdot s(n)$. Treating $n$ as continuous, find the clutch size maximizing expected survivors and comment on the integer answer.

Solution $W(n) = n\left(1 - \dfrac{n}{20}\right) = n - \dfrac{n^2}{20}$. $W'(n) = 1 - \dfrac{n}{10} = 0 \implies n = 10$. $W'' = -\tfrac{1}{10} < 0$, a max. $W(10) = 10 - 5 = 5$ expected survivors. Here $n = 10$ is already an integer, so no rounding is needed; otherwise you would compare $W$ at the two nearest integers. This is **Lack's clutch-size principle**: selection favors the brood size maximizing surviving young, not the largest possible brood — an optimization fixed point evolution discovers (§10.12). (§10.12)

F4. ⭐⭐ A drug's blood concentration after a dose is $C(t) = 5(e^{-0.2t} - e^{-0.6t})$ (mg/L) at time $t$ hours. Find the time of peak concentration.

Solution $C'(t) = 5(-0.2e^{-0.2t} + 0.6e^{-0.6t}) = 0 \implies 0.6e^{-0.6t} = 0.2e^{-0.2t} \implies 3 = e^{0.4t} \implies t = \dfrac{\ln 3}{0.4} \approx 2.75$ h. For small $t$, $C' > 0$ (rising); for large $t$, $C' < 0$ (falling), so this is the peak. Peak time $\boxed{t = \tfrac{\ln 3}{0.4} \approx 2.75 \text{ h}}$. (§10.12, §10.15)

Part G — Synthesis and Verification (§10.13–§10.15)

G1. ⭐⭐ Among all rectangles with diagonal of fixed length $d$, which has maximum area? Mind the max-vs-min distinction.

Solution Sides $a, b$ with $a^2 + b^2 = d^2 \implies b^2 = d^2 - a^2$. Maximize $A^2 = a^2 b^2 = a^2(d^2 - a^2) = a^2 d^2 - a^4$ (maximizing $A^2$ maximizes $A$). $\dfrac{d(A^2)}{da} = 2a d^2 - 4a^3 = 2a(d^2 - 2a^2) = 0 \implies a^2 = d^2/2$, so $a = d/\sqrt2 = b$: a **square**. Second derivative $2d^2 - 12a^2 = 2d^2 - 6d^2 = -4d^2 < 0$, a max. The square maximizes area for fixed diagonal — contrast §10.13, where the square *minimizes* the diagonal for fixed perimeter. Same shape, opposite extremum: read the question carefully. (§10.13)

G2. ⭐⭐⭐ Maximum power transfer (§10.14). A source with EMF $V = 12$ V and internal resistance $\rho = 4\ \Omega$ drives a load $R$. Power delivered is $P(R) = \dfrac{V^2 R}{(\rho + R)^2}$. Find the $R$ maximizing power and the power delivered.

Solution $P'(R) = \dfrac{V^2(\rho - R)}{(\rho + R)^3}$ (quotient rule, §10.14) $= 0 \implies R = \rho = 4\ \Omega$. For $R < \rho$, $P' > 0$; for $R > \rho$, $P' < 0$ — rises then falls, a global max. $P(4) = \dfrac{144 \cdot 4}{8^2} = \dfrac{576}{64} = 9$ W. **Impedance matching**: max power when load equals source resistance. (§10.14)

G3. ⭐⭐⭐ A student maximizing $f(x) = 12x - x^3$ on $[0, 3]$ finds the critical point and stops there, reporting it as the max. Find the true global max and min, and identify the omission.

Solution $f'(x) = 12 - 3x^2 = 0 \implies x^2 = 4 \implies x = 2$ (in $[0,3]$). The student reports $f(2) = 24 - 8 = 16$. But the **endpoints must be checked** (§10.4 pitfall): $f(0) = 0$, $f(3) = 36 - 27 = 9$. Comparing $\{0, 16, 9\}$: global **max** $= 16$ at $x = 2$ (the student was lucky — it *is* the max), global **min** $= 0$ at $x = 0$. The omission was skipping the endpoints; the student got the max by chance but would have missed the min. (§10.4)

G4. ⭐⭐⭐⭐ General fixed-fencing pen against a wall. A pen against a wall (three sides fenced) uses fixed fencing length $L$. Prove the maximum-area pen always has its wall-parallel side equal to $L/2$ and each perpendicular side equal to $L/4$, regardless of $L$, and find the maximum area.

Solution Perpendicular sides $x$ each, parallel side $y$: constraint $2x + y = L \implies y = L - 2x$. Area $A(x) = x(L - 2x) = Lx - 2x^2$, domain $[0, L/2]$. $A'(x) = L - 4x = 0 \implies x = L/4$, then $y = L - L/2 = L/2$. $A'' = -4 < 0$, a max; endpoints give $A = 0$. So **for every $L$**, perpendicular sides $= L/4$ and parallel side $= L/2 = 2x$, with maximum area $A = \tfrac{L}{4}\cdot\tfrac{L}{2} = \tfrac{L^2}{8}$. This is the §10.5 result generalized: the structure is scale-free. (§10.5, §10.15)

Reflection

Look back over the parts and notice how little the calculus changed: in nearly every problem you set the first derivative to zero, solved, and confirmed with a sign or second-derivative argument. What changed was the translation — fences, cans, rivers, profit, foragers — and the domain bookkeeping: closed intervals demand endpoint checks (§10.4), open intervals demand boundary limits (§10.6), and a single critical point with constant concavity settles things outright (§10.15). The four-star problems (B8, E5, F3, G4) were harder only because they asked for a general formula or a careful interpretation, not because their derivatives were exotic. Master the setup, respect the domain, and justify the global claim — that is the whole craft of single-variable optimization, and it is the foundation the multivariable methods of Chapter 31 generalize.