# Minimizing the surface area of a rectangular prism

Why is it that a rectangular prism with known sum of all side lengths that is a cube has the greatest surface area while a square based prism with known volume is a cube that has the least surface area?

Let the side-lengths of a rectangular prism be $a, b, c$. Suppose that $a+b+c = P$. Now we have $$(a-b)^2+(b-c)^2+(c-a)^2 \geq 0.$$ Expanding yields $$2a^2+2b^2+2c^2-2ab-2bc-2ca \geq 0 \Rightarrow a^2+b^2+c^2 \geq ab+bc+ca.$$ Now if we add $2ab+2bc+2ca$ to both sides we get $$a^2+b^2+c^2+2ab+2bc+2ca \geq 3ab+3bc+3ca,$$ which gives you $$P^2 = (a+b+c)^2 \geq 3(ab+bc+ca) \Rightarrow 2(ab+bc+ca) \leq \dfrac{2P^2}{3}.$$
The inequality is equality when $a=b=c$, or when the prism is a cube. This is when the surface area, $2(ab+bc+ca)$ is minimized.
If the volume $V=abc$ is fixed, then by the Arithmetic Mean Geometric Mean Inequality: $$\dfrac{ab+bc+ca}{3} \geq \sqrt[3]{a^2b^2c^2} = \sqrt[3]{V^2},$$ so $$2(ab+bc+ca) \geq 6\sqrt[3]{V^2}.$$ The equality case of AM-GM is $ab=bc=ca$ which implies $a=b=c$, so again a cube minimizes the surface area.
• Sorry, it is indeed 216, I'm not sure what I was thinking when I wrote 54. To answer your question of why one is an upper bound and one is a lower bound, that is simply the nature of the quantities $(a+b+c)^2$, $ab+bc+ca$, and $abc$ when compared with each other; we always have $(a+b+c)^2 \geq 3(ab+bc+ca) \geq 9\sqrt[3]{a^2b^2c^2}$. May 29, 2017 at 0:31