Given:
The diameter of the right cylinder is 2x cm.
The total surface area is 96 cm cube.
The radius is calculated as,

The total surface area is,

Volume is,

Now, differentiate with respect to x,

Now,
![\begin{gathered} \frac{dV}{dx}=0 \\ 84-3\pi(x^2)=0 \\ x^2=\frac{16}{\pi} \\ x=\sqrt[]{\frac{16}{\pi}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7BdV%7D%7Bdx%7D%3D0%20%5C%5C%2084-3%5Cpi%28x%5E2%29%3D0%20%5C%5C%20x%5E2%3D%5Cfrac%7B16%7D%7B%5Cpi%7D%20%5C%5C%20x%3D%5Csqrt%5B%5D%7B%5Cfrac%7B16%7D%7B%5Cpi%7D%7D%20%5Cend%7Bgathered%7D)
Now, differentiate (1) with respect to x again,
![\begin{gathered} \frac{d^2V}{dx^2}=\frac{d}{dx}(84-3\pi(x^2)) \\ =-6\pi x \\ At\text{ x=}\sqrt[]{\frac{16}{\pi}} \\ \frac{d^2V}{dx^2}=-6\pi\sqrt[]{\frac{16}{\pi}}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7Bd%5E2V%7D%7Bdx%5E2%7D%3D%5Cfrac%7Bd%7D%7Bdx%7D%2884-3%5Cpi%28x%5E2%29%29%20%5C%5C%20%3D-6%5Cpi%20x%20%5C%5C%20At%5Ctext%7B%20x%3D%7D%5Csqrt%5B%5D%7B%5Cfrac%7B16%7D%7B%5Cpi%7D%7D%20%5C%5C%20%5Cfrac%7Bd%5E2V%7D%7Bdx%5E2%7D%3D-6%5Cpi%5Csqrt%5B%5D%7B%5Cfrac%7B16%7D%7B%5Cpi%7D%7D%3C0%20%5C%5C%20%20%5Cend%7Bgathered%7D)
Since, the double derivative is negative.
![So,\text{ the volume is maximum at }\sqrt[]{\frac{16}{\pi}}](https://tex.z-dn.net/?f=So%2C%5Ctext%7B%20the%20volume%20is%20maximum%20at%20%7D%5Csqrt%5B%5D%7B%5Cfrac%7B16%7D%7B%5Cpi%7D%7D)
So, the volume becomes,
![\begin{gathered} V=\pi(x^2)h \\ V=\pi(\sqrt[]{\frac{16}{\pi}})^2h \\ V=\frac{16h}{\pi} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20V%3D%5Cpi%28x%5E2%29h%20%5C%5C%20V%3D%5Cpi%28%5Csqrt%5B%5D%7B%5Cfrac%7B16%7D%7B%5Cpi%7D%7D%29%5E2h%20%5C%5C%20V%3D%5Cfrac%7B16h%7D%7B%5Cpi%7D%20%5Cend%7Bgathered%7D)
Answer: maximum volume of the cylinder is,