1
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ you say that if sum of side lengths is known, then minimum possible surface area is when the rectangular prism is a cube, right? But for example, when the sum of all the edges of a rectangular prism is 72, the maximum surface area is when it is a cube, right? Please explain this as i am only in grade 8 and only somewhat understood the 1st part of your answer. $\endgroup$ Commented May 27, 2017 at 17:31
  • $\begingroup$ There are twelve edges of a rectangular prism, so if the sum of all of them is 72 and it is a cube then each edge should be 6. That makes the surface area 54. Is that what you're asking? $\endgroup$ Commented May 27, 2017 at 20:11
  • $\begingroup$ Wouldn't a cube with s = 6 have a surface area of 216? What I am asking is that you say that a rectangular prism with a known sum of sides has the least surface area when it is a cube. But, a cube with 6 has the maximum possible surface area. Why is it that a rectangular prism with known sum of sides has largest possible surface area when it is cube, but a square based pyramid with known volume has last possible surface area when it is a cube? $\endgroup$ Commented May 28, 2017 at 21:12
  • $\begingroup$ 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 $(a+b+c)^2 \geq 3(ab+bc+ca) \geq 9\sqrt[3]{a^2b^2c^2}$. $\endgroup$ Commented May 29, 2017 at 0:31
  • $\begingroup$ Ooohhhhh! Thanks a lot, I understand now, it's because of the inequality. Thanks! $\endgroup$ Commented May 30, 2017 at 1:24

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.