A right circular cylinder is inscribed in a sphere with diameter 4cm as shown. If the cylinder is open at both ends, find the largest possible surface area of the cylinder

Answer:[tex]8\pi\text{ square cm}[/tex]Step-by-step explanation:Since, we know that,The surface area of a cylinder having both ends in both sides,[tex]S=2\pi rh[/tex]Where,r = radius,h = height,Given, Diameter of the sphere = 4 cm,So, by using Pythagoras theorem,[tex]4^2 = (2r)^2 + h^2[/tex]   ( see in the below diagram ),[tex]16 = 4r^2 + h^2[/tex][tex]16 - 4r^2 = h^2[/tex][tex]\implies h=\sqrt{16-4r^2}[/tex]Thus, the surface area of the cylinder,[tex]S=2\pi r(\sqrt{16-4r^2})[/tex]Differentiating with respect to r,[tex]\frac{dS}{dr}=2\pi(r\times \frac{1}{2\sqrt{16-4r^2}}\times -8r + \sqrt{16-4r^2})[/tex][tex]=2\pi(\frac{-4r^2+16-4r^2}{\sqrt{16-4r^2}})[/tex][tex]=2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})[/tex]Again differentiating with respect to r,[tex]\frac{d^2S}{dt^2}=2\pi(\frac{\sqrt{16-4r^2}\times -16r + (-8r^2+16)\times \frac{1}{2\sqrt{16-4r^2}}\times -8r}{16-4r^2})[/tex]For maximum or minimum,[tex]\frac{dS}{dt}=0[/tex][tex]2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})=0[/tex][tex]-8r^2 + 16 = 0[/tex][tex]8r^2 = 16[/tex][tex]r^2 = 2[/tex][tex]\implies r = \sqrt{2}[/tex]Since, for r = √2,[tex]\frac{d^2S}{dt^2}=negative[/tex]Hence, the surface area is maximum if r = √2,And, maximum surface area,[tex]S = 2\pi (\sqrt{2})(\sqrt{16-8})[/tex][tex]=2\pi (\sqrt{2})(\sqrt{8})[/tex][tex]=2\pi \sqrt{16}[/tex][tex]=8\pi\text{ square cm}[/tex]