Question

Suppose you are given the task of constructing a soup can that holds 400mL of water (1 mL = 1 cm3 ). The material for the base and lid of the can costs $.01/100cm2 , and the material for the side of the can costs $.02/100cm2 . (a) (6 pts) Find the dimensions of a can that would minimize the cost to produce it.

Answer 1

Answer:

Dimensions:

$r = ~^3\sqrt{\frac{400}{\pi}}$

$h = ~^3\sqrt{\frac{400}{\pi}}$

Step-by-step explanation:

Volume is given by  

$\text{Volume of cylinder} = \pi r^2h = 400~cm^3\\\\h = \displaystyle\frac{400}{\pi r^2}$

where r is the radius of can and h is the height of the can.

Area of lid and base = $\pi r^2 + \pi r^2 = 2\pi r^2$

Area of curved surface = $2\pi rh$

Cost function :

$C(r) = \displaystyle\frac{0.01}{100}2\pi r^2 + \frac{0.02}{100}2\pi rh\\\\= \frac{0.02}{100}\pi r^2 + \frac{0.04}{100}\pi r \frac{400}{\pi r^2}\\\\= \frac{0.02}{100}\pi r^2 + \frac{0.16}{r}$

First, we differentiate C(r) with respect to r, to get,

$\displaystyle\frac{d(C(r))}{dr} = \displaystyle\frac{0.04}{100}\pi r - \frac{0.16}{r^2}$

Equating the first derivative to zero, we get,

$\displaystyle\frac{d(C(r))}{dr} = 0\\\\\frac{0.04}{100}\pi r - \frac{0.16}{r^2} = 0$

Solving, we get,

$\displaystyle\frac{0.04}{100}\pi r - \frac{0.16}{r^2} = 0\\\\\frac{0.04}{100}\pi r = \frac{0.16}{r^2}\\\\r^3 = \frac{0.16\times 100}{0.04\pi}\\\\r =~^3\sqrt{\frac{400}{\pi}} = \bigg(\frac{400}{\pi}\bigg)^{\frac{1}{3}}$

Again differentiation C(r), with respect to r, we get,

$\displaystyle\frac{d^2(C(r))}{dr^2} =\displaystyle\frac{0.04}{100}\pi+\frac{0.32}{r^3}$

At r = $\bigg(\displaystyle\frac{400}{\pi}\bigg)^{\frac{1}{3}}$,

$\displaystyle\frac{d^2(C(r))}{dr^2} > 0$

Thus, by double derivative test, the minima occurs for C(r) at  r = $\bigg(\displaystyle\frac{400}{\pi}\bigg)^{\frac{1}{3}}$.

Dimensions:

$r = \bigg(\displaystyle\frac{400}{\pi}\bigg)^{\frac{1}{3}}$

$h = \displaystyle\frac{400}{\pi r^2}\\\\h = \frac{400}{\pi (\frac{400}{\pi})^{\frac{2}{3}}} = ~^3\sqrt{\frac{400}{\pi}}$