Optimization Practice

These questions follow Optimization — warm-ups first, then the boxes, then models drawn from the world. For every question: name the variable, build the function, state the domain the story allows, and audit your candidate before you crown it.

Warming up

1. Two numbers have a sum of 30. Using calculus, determine the pair with the greatest product.
2. A farmer has 1200 m of fencing for a rectangular field along a straight river — no fence needed on the river side. What dimensions give the largest area?

Answer 1

Let one number be $x$; the other is $30-x$. Maximise $P(x)=x(30-x)=30x-x^2$. Then $P^{\prime }(x)=30-2x$, zero at $x=15$; $P^{\prime }$ is positive before and negative after, so this is a maximum. The pair is 15 and 15, with product 225. The estimate you made before differentiating — “probably the middle” — just got certified.

Answer 2

Let $x$ be each side perpendicular to the river; the parallel side uses what remains, $1200-2x$. Area $A(x)=x(1200-2x)=1200x-2x^2$ on $0<x<600$. $A^{\prime }(x)=1200-4x$, zero at $x=300$, with $A^{\prime }$ changing $+$ to $-$: a maximum. Dimensions $300$ m by $600$ m, area $180000$ m². Not a square — the river is doing the work of one fence, and the optimum leans into the free side.

Boxes and packages

3. An open-top box is made from a 24 cm by 24 cm sheet by cutting equal squares of side $x$ from the corners and folding up the sides. Determine the value of $x$ that maximises the volume, and the maximum volume.
4. An open-top box with a square base must hold 4000 cm³. What dimensions minimise the material used?

Answer 3

$V(x)=x(24-2x{)}^2$ on $0<x<12$. Product and chain rules: $\begin{array}{rl}{V}^{\prime }(x)& =(24-2x{)}^2+x\cdot 2(24-2x)(-2)\\ & =(24-2x)(24-6x)\end{array}$ Zero at $x=12$ (the domain’s edge — a flattened box of volume zero) and $x=4$. Sign of $V^{\prime }$ around 4: positive then negative — a maximum. $V(4)=4(16{)}^2=1024$ cm³. This is The Box Problem settled in general: every group’s fold was a point on this curve, and $x=4$ is its summit.

Answer 4

Base side $x$, height $h$, with $x^{2}h=4000$, so $h=\frac{4000}{x^2}$. Material (base plus four sides): $S(x)=x^2+4xh=x^2+\frac{16000}{x},x>0$ $S^{\prime }(x)=2x-\frac{16000}{x^2}$, zero when $x^3=8000$, so $x=20$. Audit: $S^{\prime \prime }(x)=2+\frac{32000}{x^3}>0$ — concave up, a genuine minimum. Dimensions: base 20 cm by 20 cm, height $h=\frac{4000}{400}=10$ cm, using $S(20)=400+800=1200$ cm². The height is half the base side — a shape worth remembering for The Packaging Brief.

Models from the world

5. The number of daily bus riders is modelled by $1200(1.15{)}^{-x}$, where $x$ is the fare in dollars. What fare maximises the total revenue?
6. A foraging bird gains $E=\frac{3000t}{t+4}$ joules of food energy by spending $t$ minutes in a berry patch, and takes 2 minutes to fly to each new patch. How long should it stay in a patch to maximise its average rate of energy gain?

Answer 5

Revenue is fare times riders: $R(x)=1200x(1.15{)}^{-x}$. Product rule, with $\frac{d}{dx}(1.15{)}^{-x}=-(1.15{)}^{-x}\ln 1.15$: $R^{\prime }(x)=1200(1.15{)}^{-x}\left(1-x\ln 1.15\right)$ The exponential factor is never zero, so $R^{\prime }(x)=0$ exactly when $x=\frac{1}{\ln 1.15}\approx 7.16$. $R^{\prime }$ is positive before and negative after: a maximum. A fare of about $7.16 — call it $7.15 at the fare box — yields revenue near $R(7.16)\approx \text{textdollar}3159$ per day. Cheaper fares carry more riders for too little each; dearer fares charge fewer riders too much.

Answer 6

Average rate of gain = energy divided by total time, foraging plus travel: $R(t)=\frac{3000t}{(t+4)(t+2)}=\frac{3000t}{t^2+6t+8}$ Writing this as $3000t(t^2+6t+8{)}^{-1}$ and differentiating: $\begin{array}{rl}{R}^{\prime }(t)& =\frac{3000(t^2+6t+8)-3000t(2t+6)}{(t^2+6t+8{)}^2}\\ & =\frac{3000(8-t^2)}{(t^2+6t+8{)}^2}\end{array}$ Zero at $t=\sqrt{8}=2\sqrt{2}\approx 2.8$ minutes, with $R^{\prime }$ positive before and negative after: a maximum. The bird should stay about 2.8 minutes — leave while berries remain, because the hidden ones cost more time than a fresh patch does. Compare strategies, as the curriculum asks: a table of $R(t)$ and a graph both point to the same summit the algebra found.