§3.6 Applied optimization

Introduction

More applied optimization problems

Summary

A parcel with square end $x$ and length $y$ obeys girth plus length $4x + y = 108$. Eliminating $y$, the volume function is

Solving $V'(x) = 216x - 12x^2 = 0$ in the postal problem gives $x = 18$ (with $y = 36$), and the largest possible volume equals

A cylindrical can has radius $r$, height $h$, lids costing $0.027$ and side costing $0.015$ per square inch. The volume and cost expressions are

A cylindrical can must hold $16$ cubic inches, with lids costing more than the side. Before differentiating the cost, the necessary first move is

For the can with $C(r) = 0.054\pi r^2 + \frac{0.48}{r}$, the only critical number is $r = \sqrt[3]{\frac{0.48}{0.108\pi}} \approx 1.123$, and $C''(r) \gt 0$ everywhere. The minimum cost is about

A square-bottomed box with top and bottom at \3.00$per square foot and sides at$\$1.50$ has budget $15 = 6x^2 + 6xh$, so $h = \frac{15 - 6x^2}{6x}$ and $V(x) = \frac{15}{6}x - x^3$ on $0 \lt x \lt \sqrt{\frac{15}{6}}$. The critical number of $V$ is

For the box with $V(x) = \frac{15}{6}x - x^3$ and critical number $x = \sqrt{\frac{15}{18}}$, compute the maximum volume, rounded to three decimals.

A farmer encloses a rectangle and adds three inner fences parallel to the sides, using $2x + 5y = 7500$ feet total. With $A(y) = 3750y - 2.5y^2$, the critical number is $y = 750$, and each of the four pens has area

Poles of heights $60$ and $80$ feet stand $100$ feet apart, and a cable runs from each top to a ground point $x$ feet from the first pole. With $L(x) = \sqrt{x^2+3600} + \sqrt{(100-x)^2+6400}$, the derivative vanishes near $x \approx 42.857$, and the candidate values are $L(0) \approx 188.06$, $L(100) \approx 196.62$, $L(42.857) \approx 172.05$. The minimum cable length is about

A cylindrical tank with hemispherical ends holds $1000$ cubic feet, with ends at \5$and barrel at$\$2$ per square foot. After eliminating the barrel length, the cost becomes $C(r) = \frac{4000}{r} + \frac{44\pi}{3}r^2$, whose only critical number is $r = \sqrt[3]{\frac{1500}{11\pi}} \approx 3.514$. The minimum cost is about

In the postal problem with $V(x) = 108x^2 - 4x^3$, factor $V'(x) = 12x(18 - x)$ and give the positive critical number that maximizes volume.

For the can with $C(r) = 0.054\pi r^2 + \frac{0.48}{r}$, the minimum occurs at $r \approx 1.123$. Compute the minimum cost $C(1.123)$, rounded to three decimals.

In the can problem, once the volume is fixed the cost function must be rewritten in terms of the radius alone before differentiating to find the minimum.