[Math] Find the dimensions of a cylinder of given volume V if its surface area is a minimum.

derivatives

The following is the question :

Find the dimensions of a cylinder of given volume V if its surface area is a minimum.

The cylinder has a closed top and bottom.

2 formula :

(1) $V=r^2\pi h$

(2) $A=2r\pi h+2r^2\pi$ -> $A=2r\pi \left(h+r\right)$

I cannot find the equation for differentiation

How to find $A'$? Hints?

Thank your for your attention

Best Answer

The volume is given, so that is a constant.

The volume constraint gives $h=\frac{V}{\pi r^2}$, from which we get $A(r) = 2 \pi r (r+\frac{V}{\pi r^2})$.

We see that $\lim_{r \downarrow 0} A(r) = \infty$ and $\lim_{r \to \infty} A(r) = \infty$, hence $A$ has a minimum.

Differentiate $A$ and set the derivative to zero. Solve for $r_0$. Then compute the corresponding $h_0$.

Details:

We have $A'(r) =-\frac{2\,\left( V-2\,\pi \,{r}^{3}\right) }{{r}^{2}}$, hence there is exactly one point $r_0$ for which $A'(r_0) = 0$, hence this must be he minimum. This gives$r_0 = \sqrt[3]{{V \over 2 \pi}}$. Then $h_0 = \frac{V}{\pi r_0^2} = \sqrt[3]{{4V \over \pi}}$.

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