Show that the height of a cylinder, which is open at the top,having a given surface area and greatest volume,is equal to the radius of its base.

 

 

 

 
 
 
 
 

Answers (1)

Let R be the radius
H be the height
V be the volume
s be the total surface area

V= \pi R^{2}H
s= \pi R^{2}+2\pi RH
H= \frac{s-\pi R^{2}}{2\pi R}----(1)
substititing value of H in V
V= \frac{1}{2}\left ( sR-\pi R^{3} \right )\rightarrow differentiating\: equation:
\frac{dV}{dR}= \frac{1}{2}\left ( s-3\pi R^{2} \right )
\frac{dV}{dR}= 0
\Rightarrow \frac{1}{2}\left ( s-3\pi R^{2} \right )= 0---(2)
R= \sqrt{\frac{s}{3\pi }}
R= \sqrt{\frac{s}{3\pi }}---(3)
Differentiating equation (2) again
\frac{d^{2}V}{dR^{2}}= \frac{1}{2}\left ( 0-6\pi R \right )\Rightarrow -3\pi R
V is greatest value R= \sqrt{\frac{s}{3\pi }}  from equation (3)
H= \frac{s-\pi \times \frac{s}{3\pi }}{2\pi \sqrt{\frac{3}{3\pi }}}   [ substituting equation (1)   R= \sqrt{\frac{s}{3\pi }} ]
H= \frac{\frac{2s}{3}}{2\sqrt{\frac{\pi s}{3}}}= \sqrt{\frac{s}{3\pi }}=R

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