A cylinder is inscribed in a given sphere of radius a. Find the dimension of the cylinder if its lateral surface area is maximum.
Solution:
The lateral surface area of the cylinder is \(A=2\pi r_c h\):
Since the radius of the sphere is a constant, "\(a\)", rewrite the lateral surface area of the cylinder as a function of the its height or its radius with the radius of the sphere (\(a\)).
\((\frac{h}{2})^2 +(r_c)^2=a^2\)
\(r_c = \frac{1}{2} \sqrt{4a^2-h^2}\)
\(A=2\pi r_c h\)
\(A=2\pi (\frac{1}{2} \sqrt{4a^2-h^2}) h\)
\(A=\pi (\sqrt{4a^2-h^2}) \sqrt{h^2}\)
\(A=\pi \sqrt{4h^2a^2-h^4}\): The lateral surface area of the cylinder as a function of \(h\) and \(a\)
Maximize the lateral surface area \(A\).
\(\frac{dA}{dh}=\pi \frac{1}{2}\frac{8ha^2-4h^3}{\sqrt{4h^2a^2-h^4}}\)
\(0=\pi \frac{1}{2}\frac{8ha^2-4h^3}{\sqrt{4h^2a^2-h^4}}\)
\(0=8ha^2-4h^3\)
\(h^2=2a^2\)
\(h=a\sqrt{2}\)
\(r_c=a\frac{\sqrt{2}}{2}\)
Thus the dimension of the cylinder (at maximum lateral surface area) inside a sphere of radius \(a\) is:
\(h=a\sqrt{2}\)
\(r_c=a\frac{\sqrt{2}}{2}\)