Question

Of all the closed right circular cylindrical cans of volume $128\pi c{m}^3$, find the dimensions of the can which has minimum surface area.

Answer 1

Given: $V=128\pi =\pi r^{2}h\Rightarrow h=\frac{128}{r^2}$

Minimize: $S=2\pi rh+2\pi r^2$

Put $h=\frac{128}{r^2}$

Therefore, $S\left(r\right)=\frac{256\pi }{r}+2\pi r^2\Rightarrow S^{\prime }\left(r\right)=-\frac{256\pi }{r^2}+4\pi r$

$S^{\prime \prime }\left(r\right)=-\frac{2\times 256\pi }{r^3}+4\pi$

For $S^{\prime }\left(r\right)=0,\Rightarrow -\frac{256\pi }{r^2}+4\pi r\Rightarrow r=4$ cm

Now, $S^{\prime \prime }\left(4\right)>0$ at r = 4 is point of minima.

$h=\frac{128\pi }{\pi \times 4\times 4}=8$ cm

Therefore, minimum surface area the can should have a radius of $4$ cm and height of $8$ cm.