A right circular cylinder is inscribed in a sphere of radius $r = 5 \mathrm {~cm}$. Find the largest possible surface area of such a cylinder. Round the result to the nearest hundredth.
A) $157.08 \mathrm {~cm} ^ { 2 }$
B) $177.08 \mathrm {~cm} ^ { 2 }$
C) $187.08 \mathrm {~cm} ^ { 2 }$
D) $167.08 \mathrm {~cm} ^ { 2 }$
E) $197.08 \mathrm {~cm} ^ { 2 }$

Question

Quizplus · Accepted Answer

Let the cylinder have height $h$ and radius $r$. Then, the diameter of the sphere is the diagonal of the cylinder's rectangular cross-section, or $\sqrt{h^2 + 4r^2}$. By the Pythagorean Theorem, $h^2 + 4r^2 = (2r)^2$, or $h^2 = 0$. Thus, $h=0$ for the cylinder with the maximum surface area, which means the cylinder is a degenerate cylinder and coincides with the sphere. The surface area of the cylinder is then $2\pi rh + 2\pi r^2 = 2\pi\cdot(5)(r) + 2\pi r^2 = 10\pi r + 2\pi r^2$. Using $r=5$, we get a maximum surface area of $10\pi(5) + 2\pi(5^2) = \boxed{157.08}$ cm$^2$.

A Right Circular Cylinder Is Inscribed in a Sphere of Radius