Answer:
![8\pi\text{ square cm}](https://tex.z-dn.net/?f=8%5Cpi%5Ctext%7B%20square%20cm%7D)
Step-by-step explanation:
Since, we know that,
The surface area of a cylinder having both ends in both sides,
![S=2\pi rh](https://tex.z-dn.net/?f=S%3D2%5Cpi%20rh)
Where,
r = radius,
h = height,
Given,
Diameter of the sphere = 4 cm,
So, by using Pythagoras theorem,
( see in the below diagram ),
![16 = 4r^2 + h^2](https://tex.z-dn.net/?f=16%20%3D%204r%5E2%20%2B%20h%5E2)
![16 - 4r^2 = h^2](https://tex.z-dn.net/?f=16%20-%204r%5E2%20%3D%20h%5E2)
![\implies h=\sqrt{16-4r^2}](https://tex.z-dn.net/?f=%5Cimplies%20h%3D%5Csqrt%7B16-4r%5E2%7D)
Thus, the surface area of the cylinder,
![S=2\pi r(\sqrt{16-4r^2})](https://tex.z-dn.net/?f=S%3D2%5Cpi%20r%28%5Csqrt%7B16-4r%5E2%7D%29)
Differentiating with respect to r,
![\frac{dS}{dr}=2\pi(r\times \frac{1}{2\sqrt{16-4r^2}}\times -8r + \sqrt{16-4r^2})](https://tex.z-dn.net/?f=%5Cfrac%7BdS%7D%7Bdr%7D%3D2%5Cpi%28r%5Ctimes%20%5Cfrac%7B1%7D%7B2%5Csqrt%7B16-4r%5E2%7D%7D%5Ctimes%20-8r%20%2B%20%5Csqrt%7B16-4r%5E2%7D%29)
![=2\pi(\frac{-4r^2+16-4r^2}{\sqrt{16-4r^2}})](https://tex.z-dn.net/?f=%3D2%5Cpi%28%5Cfrac%7B-4r%5E2%2B16-4r%5E2%7D%7B%5Csqrt%7B16-4r%5E2%7D%7D%29)
![=2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})](https://tex.z-dn.net/?f=%3D2%5Cpi%28%5Cfrac%7B-8r%5E2%2B16%7D%7B%5Csqrt%7B16-4r%5E2%7D%7D%29)
Again differentiating with respect to r,
![\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})](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E2S%7D%7Bdt%5E2%7D%3D2%5Cpi%28%5Cfrac%7B%5Csqrt%7B16-4r%5E2%7D%5Ctimes%20-16r%20%2B%20%28-8r%5E2%2B16%29%5Ctimes%20%5Cfrac%7B1%7D%7B2%5Csqrt%7B16-4r%5E2%7D%7D%5Ctimes%20-8r%7D%7B16-4r%5E2%7D%29)
For maximum or minimum,
![\frac{dS}{dt}=0](https://tex.z-dn.net/?f=%5Cfrac%7BdS%7D%7Bdt%7D%3D0)
![2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})=0](https://tex.z-dn.net/?f=2%5Cpi%28%5Cfrac%7B-8r%5E2%2B16%7D%7B%5Csqrt%7B16-4r%5E2%7D%7D%29%3D0)
![-8r^2 + 16 = 0](https://tex.z-dn.net/?f=-8r%5E2%20%2B%2016%20%3D%200)
![8r^2 = 16](https://tex.z-dn.net/?f=8r%5E2%20%3D%2016)
![r^2 = 2](https://tex.z-dn.net/?f=r%5E2%20%3D%202)
![\implies r = \sqrt{2}](https://tex.z-dn.net/?f=%5Cimplies%20r%20%3D%20%5Csqrt%7B2%7D)
Since, for r = √2,
![\frac{d^2S}{dt^2}=negative](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E2S%7D%7Bdt%5E2%7D%3Dnegative)
Hence, the surface area is maximum if r = √2,
And, maximum surface area,
![S = 2\pi (\sqrt{2})(\sqrt{16-8})](https://tex.z-dn.net/?f=S%20%3D%202%5Cpi%20%28%5Csqrt%7B2%7D%29%28%5Csqrt%7B16-8%7D%29)
![=2\pi (\sqrt{2})(\sqrt{8})](https://tex.z-dn.net/?f=%3D2%5Cpi%20%28%5Csqrt%7B2%7D%29%28%5Csqrt%7B8%7D%29)
![=2\pi \sqrt{16}](https://tex.z-dn.net/?f=%3D2%5Cpi%20%5Csqrt%7B16%7D)
![=8\pi\text{ square cm}](https://tex.z-dn.net/?f=%3D8%5Cpi%5Ctext%7B%20square%20cm%7D)