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# 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.

Toolbox:
• Surface Area $S=2\pi r^2+2\pi rh$
• $h=\large\frac{s-2\pi r^2}{2\pi r}$
• Volume $V=\pi r^2h$
Step 1:
Let $S$ be the given surface area of the closed cylinder whose radius is $r$ and height $h$.
Let $V$ be its volume.
Surface Area $S=2\pi r^2+2\pi rh$
$h=\large\frac{S-2\pi r^2}{2\pi r}$----(1)
Volume $V=\pi r^2h$
Substitute the value of h in the above equation
$\qquad\qquad=\pi r^2\bigg[\large\frac{S-2\pi r^2}{2\pi r}\bigg]$
$\qquad\qquad=\large\frac{1}{2}$$r(S-2\pi r) \qquad\qquad=\large\frac{1}{2}$$[Sr-2\pi r^3]$
Differentiating with respect to r we get
$\large\frac{dV}{dx}=\frac{1}{2}$$[S-6\pi r^2] Step 2: For maxima and minima \large\frac{dV}{dx}$$=0$
$S-6\pi r^2=0$
$S=6\pi r^2$
From (1)
$h=\large\frac{S-2\pi r^2}{2\pi r}$
Substitute the value os S in the above equation
$h=\large\frac{6\pi r^2-2\pi r^2}{2\pi r}$
$\;\;=\large\frac{4\pi r^2}{2\pi r}$
$\;\;=2r$
$h=2r$
Step 3:
On double differentiation of V we get,
$\large\frac{d^2V}{dr^2}=\frac{1}{2}$$(-12\pi r)$
$\qquad=-6\pi r$
$\qquad=-ve$
$\therefore V$ is maximum
Thus volume is maximum when $h=2r$
(i.e)when height of cylinder =diameter of the base.
edited Aug 19, 2013