A closed cylindrical can is to be made so that its volume is $52cm^3$. Find its dimensions such that the surface area is to be minimum.
My attempt
Let $x$ be the radius $y$ be the height of the cylinder. Then
$$\textrm {Volume}=\pi x^2y$$
$$52=\pi x^2y$$
$$y=\dfrac {52}{\pi x^2}$$
Let $f(x)=\dfrac {52}{\pi x^2}$
$$f'(x)=-\dfrac {104}{x^2}$$
$f'(x)=0$ gives
$$\dfrac {104}{x^2}=0$$
Best Answer
The volume is $$V=\pi r^2 h $$
the surface area is $$S(r)=2\pi r^2+2\pi r h $$ $$=2\pi r^2 +2\pi r \frac {V}{\pi r^2} $$ $$=2\pi r^2+\frac {2V}{r} $$
$$S'(r)=4\pi r -\frac {2V}{r^2}=0$$
thus $$r=(\frac {V}{2\pi})^\frac13$$