Show that the right circular cylinder of given surface

Question: Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.

Solution:  Let r be the radius of the circular base and h be the height of closed right circular cylinder.

Show that the right circular cylinder of given surface

Formula for Total surface area (\mathrm{S})=2 \pi r h+2 \pi r^2

S-2\pi r^2 = 2\pi rh

\Rightarrow h= \frac{S-2\pi r^2}{2\pi r} –(i)

Volume of cylinder (V) = \pi r^2 h

V = \pi r^2(\frac{S-2\pi r^2}{2\pi r})

\Rightarrow V = \frac{1}{2}(Sr - 2\pi r^3)

Differentiating with respect r

\frac{dV}{dr} = \frac{1}{2}(S-6\pi r^2)

And \frac{d^2V}{dr^2} = \frac{1}{2}(0-12\pi r)=-6\pi r

For max and minima \frac{dV}{dr}= 0

\frac{1}{2}(S-6\pi r^2)=0

\Rightarrow S = 6\pi r^2.

Since \frac{d^2V}{dr^2} = -6\pi r<0

Hence volume of cylinder is max when S = 6\pi r^2

Value of S putting in equation (i)

h= \frac{6\pi r^2-2\pi r^2}{2\pi r}

\Rightarrow h = \frac{4\pi r^2}{2\pi r} = 2r

Height of cylinder = Diametre of cylinder

Hence volume of cylindr is max when h = 2r

 

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