# What dimensions of the rectangle will result in a cylinder of maximum volume if you consider a rectangle of perimeter 12 inches in which it forms a cylinder by revolving this rectangle about one of its edges?

Feb 5, 2015

A cylinder's volume, $V = \pi {r}^{2} h$
The width of the rectangle, which forms the circumference, $w = 2 \pi r$
graph{x+y=12 [-2.49, 25.99, -1.51, 12.74]}
The graph shows possible lengths for the sides of the rectangle if they must add up to 12 inches ($w$ is horizontal and $h$ is the vertical axis).
At what point along that curve is ${w}^{2} / \left(4 {\pi}^{2}\right) \cdot h = {r}^{2} h$ the greatest?

graph{(12x-x^2)/(4pi) [-0.89, 13.35, -1.027, 6.1]}

Since $r = \frac{w}{2 \pi} = \frac{12 - h}{2 \pi}$, you can plot a graph of the volume for different values of $h$ with the equation $y = \pi \cdot \frac{12 - x}{4 {\pi}^{2}} \cdot x = \frac{12 x - {x}^{2}}{4 \pi}$. Differentiating to find the maximum:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{12}{4 \pi} - \frac{2 x}{4 \pi} = \frac{3}{\pi} - \frac{x}{2 \pi} = 0$

$x = h = 6$

Since $h = 6$, $w$ must also equal $6$

$V = \pi {r}^{2} \cdot h = \pi \cdot {\left(\frac{6}{2 \pi}\right)}^{2} \cdot 6 = 533 \pi = 1674$