A closed cylindrical tank having a volume of 71.57 m^3 is to be constructed. If the surface area is to be minimum, what is the required diameter of the tank?

For maximum volume
$H = D$
 

 

$V = \frac{1}{4}\pi D^2 H$

$71.57 = \frac{1}{4}\pi D^3$

$D = 4.5 ~ \text{m}$

Volume of the tank
$V = \frac{1}{4}\pi D^2 H$

$71.57 = \frac{1}{4}\pi D^2 H$

$H = \dfrac{286.28}{\pi D^2}$
 

 

Surface area of the tank
$A = 2A_b + A_L$

$A = 2 \left( \frac{1}{4}\pi D^2 \right) + \pi DH$

$A = \frac{1}{2}\pi D^2 + \pi D \left( \dfrac{286.28}{\pi D^2} \right)$

$A = \frac{1}{2}\pi D^2 + \dfrac{286.28}{D}$
 

For the surface area to be minimum
$\dfrac{dA}{dD} = \pi D - \dfrac{286.28}{D^2} = 0$

$\pi D = \dfrac{286.28}{D^2}$

$D^3 = \dfrac{286.28}{\pi}$

$D = 4.5 ~ \text{m}$