How do you optimize a donut

Question

I need to find the minimum surface area of a donut by keeping the volume the same. I know that the volume of my donut is $155 \,cm^3$ and the formula for it is $2a^2\pi^2b$ ($a$ is the inner radius, $b$ is the outer radius) and the formula for the surface area is $4\pi^2ab$ - can someone help solve this?

Answer 1

We know that the volume is fixed at $V=155$ (I'll leave the units out for now). From the equation of the volume, we get $$ V = 2a^2 \pi^2 b = 155 \Rightarrow b = \frac{155}{2a^2 \pi^2} $$ Plugging this result into the equation for the surface area, we get $$ A = 4\pi^2 a b = 2 \pi^2 a \frac{155}{2a^2 \pi^2} = 2 \frac{155}{a} $$ Therefore, in order to minimise the surface area, you have to stretch $a$ to the largest value available.