Question

A cylindrical can without a top is to be made from 27cm^2

of sheet metal. Find the dimensions of the can with

the greatest volume. Be sure to formally introduce any new variables and their domains. Confirm you have a maximum

Answer 1

The area of the can will be $S =\pi R^2 + 2\pi R h,$ where R is radius of the bottom and h is the height.

The volume is $V = \pi R^2 h$ .

$S =\pi R^2 + 2\pi R h \Rightarrow h = \dfrac{S-\pi R^2}{2\pi R}, \\ V = \pi R^2\cdot \dfrac{S-\pi R^2}{2\pi R} = 0.5(SR- \pi R^3).$

Let us take the derivative with respect to R

$V' = 0.5S - 1.5\pi R^2.$

We should find a maximum, so we'll consider values for which the derivative is 0. Moreover, R should be positive.

$0.5S - 1.5\pi R^2 = 0, \;\; R = \sqrt{\dfrac{0.5S}{1.5\pi}} = \sqrt{\dfrac{S}{3\pi}} = \sqrt{\dfrac{27}{3\pi}} = \dfrac{3}{\sqrt{\pi}}\,\mathrm{cm}.$

For greater R $1.5\pi R^2$ will be greater than 0.5S, so the derivative will be negative, and for smaller values the derivative will be positive. So, $\dfrac{3}{\sqrt{\pi}}\,\mathrm{cm}$ is the point of maximum.

The corresponding volume is $0.5 \left(27\cdot\dfrac{3}{\sqrt{\pi}} - \pi \cdot \dfrac{27}{\sqrt{\pi}^3}\right) = \dfrac{27}{\sqrt{\pi}}$ .

The height will be $h = \dfrac{S-\pi R^2}{2\pi R} = \dfrac{27-\pi\frac{9}{\pi}}{2\pi \frac{3}{\sqrt{\pi}}} = \dfrac{3}{\sqrt{\pi}}.$