Question

Find the dimensions of the right-circular cylinder of greatest lateral surface area that can be
inscribed in a sphere of radius 6 inches.

Answer 1

Let $r = 6$ inches. R and h be radius and height of the cylinder.

Let R and h are related to the r by the relation,

$R = rcos\theta, h=rsin\theta$ where $\theta$ is the angle between radius of the cylinder and radius of the sphere.

Then lateral surface area is given by, $S = 2\pi rh = 2\pi R^2 sin\theta cos\theta$

For maximum or minimum, $\frac{dS}{d\theta } = 0$

then, $\frac{dS}{d\theta } = 2\pi R^2(cos^2\theta-sin^2\theta) = 0$

solving for $\theta$ , we get, $2\pi R^2(cos^2\theta-sin^2\theta) = 0$

$cos^2\theta=sin^2\theta$

$tan\theta = 1 \implies \theta = \frac{\pi}{4}$

Then, radius of the cylinder is

$r = Rcos\theta = Rcos\frac{\pi}{4} = \frac{6}{\sqrt{2}} = 3\sqrt{2} = 4.24$ inches.

Height of the cylinder is,

$h = Rsin\theta = Rsin\frac{\pi}{4} = \frac{6}{\sqrt{2}} = 3\sqrt{2} = 4.24$ inches