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My child's teacher raised a quesion in class for students who are interested to prove. The teacher says that the volume of a cube is the greatest among rectangular-faced shapes of the same perimeter and asks his students to prove this proposition.

I considered the relationship between the length of the sides of a cube and the lengths of the sides of rectangular-faced shapes in different situation. But when the calculations came down to polynomials, I couldn't proceed due to the uncertainty of the variables in the polynomials.

Can anyone please find a good way to prove the above proposition? Or is there already a proof? Thank you for your help!

Clement Yung
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Michael May
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    It should be surface area, not perimeter. If you can prove the square has the highest area among rectangles of that perimeter, you can just concentrate on two dimensions at a time. Show that if any two of the three dimensions are unequal, you can increase the volume at reduced area by making them both the average. – Ross Millikan Jul 29 '20 at 03:03
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    If sides of a rectangle have lengths $s+x$ and $s-x$, area $(s+x)(s-x)$ is maximized when $x=0$ – J. W. Tanner Jul 29 '20 at 03:10
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    For a given sum of two numbers, its multiplication is max when they are both same. You can prove that thru AM-GM, differentiation etc. That proves area of rectangle for a given perimeter is max when it is a square. Now for volume, pls consider a given surface area and prove that the volume is max when it is a cube., again using similar methods. – Math Lover Jul 29 '20 at 03:20
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    I don't think the skill that the teacher wants to teach your child here is "have your dad ask the internet for the answer". – Federico Poloni Jul 29 '20 at 16:24
  • Thanks to all who have participated in the discussion of this question. Now there's another related question. Suppose the sum of the three sides (length, width and height) happens to be a number indivisible by 3, e.g. 10, then the length of any of the cube's sides would be a repeating decimal, e.g. 1.33333... So, does that mean the cube wouldn't exist in reality? This means that the cube wouldn't stop growing in size, though it grows with an ever-reducing scale. Is that just like we couldn't have a so-called perfect circle because PI is an infinite non-repeating decimal? Thank you very much. – Michael May Jul 30 '20 at 00:02
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    Thanks, @FedericoPoloni. But please be tolerant, pal. I'm sorry I didn't mention in the thread that my child is just a primary student. The teacher had probably intended to induce his students' interest in math. I personally won't expect primary school students to solve a math problem that requires knowledge higher than what they learn at school. The teacher knew that, I guess. He might just be encouraging his students to give ti a good think, which my child did to no avail, and expected them to search the Internet for solution or ask their parents to sort out the question together with them. – Michael May Jul 30 '20 at 00:18
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    To address the question about the sum not being divisible by three. The units are arbitrary and do not determine whether or not a cube exists. For example, the sum of the edges add up to "16". Well, if the sum is 16 yards, each edge is $1.\overline{3}$ yards. But this is exactly 4 feet! So each edge is 4 feet. Anyway, even if the edges added to 12 yards, how are you going to measure one yard exactly? – mjw Aug 02 '20 at 21:27

3 Answers3

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Is elementary solutions permitted?

$$ \frac{a+b+c }{3}\geq \sqrt[3]{abc} $$

Equality i.e. maximum volume for a given sum of side lengths is when all sides are equal

acat3
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If you mean by "perimeter" the sum of the edges, then yes, the cube is the maximal rectangular parallelepiped among those with the same "perimeter".

Let the edges have lengths $(a,b,c)$.

Then the volume is $V=abc$ and the "perimeter" is $p=4(a+b+c).$

We can maximize volume while constraining the sum of the edges using Lagrange multipliers:

$$\begin{aligned} L &= abc-\lambda \left(a+b +c-\frac{p}{4}\right)\\ 0&=\frac{\partial L}{\partial a} = bc - \lambda\\ 0&=\frac{\partial L}{\partial b} = ac - \lambda\\ 0&=\frac{\partial L}{\partial c} = ab - \lambda\\ \end{aligned}$$ so that $$bc=ac=ab$$ and $$a=b=c.$$

mjw
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    Apart from using Lagrange multipliers (which I presume the OP is not supposed to know) this does not prove that that's a maximum, only a critical point. – leonbloy Jul 29 '20 at 03:22
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    Fair enough. The OP can look up Lagrange multipiers. I don't think it is any more difficult than using other elementary techniques. That's debatable. And yes, this only shows it is a critical point. To be rigorous, consider now $(a+\epsilon,b-\epsilon,c)$ (other permutations covered here by symmetry) and this is clearly a minimum, not another critical point. – mjw Jul 29 '20 at 03:25
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    In the comment above, "... clearly a $\textbf{maximum}$," should have written, not "minimum". – mjw Jul 29 '20 at 03:31
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    Please see https://mathworld.wolfram.com/LagrangeMultiplier.html – mjw Jul 29 '20 at 03:34
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    Might be an improvement to mention that the other solutions to ab=ac=bc show the other critical solutions are all 0 (a = 0 || b = 0 || c = 0), which support the maximum – Cireo Jul 29 '20 at 20:26
  • If $a=0$, for example, then either $b=0$ or $c=0$, so perhaps you mean $(a=0=b || b=0=c || c=0=a)$? – mjw Jul 29 '20 at 21:26
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Another solution to what mjw posted, this one without use of Lagrange multipliers is as following. Fix the "perimeter" $P$ such that $P=4(a+b+c)$ is constant then the volume is

$$ V=ab(P/4-a-b) $$ and take partial derivates to get $$ \frac{\partial V}{\partial a}=b(P/4-a-b)-ab=0 $$ and $$ \frac{\partial V}{\partial b}=a(P/4-a-b)-ab=0. $$ It is easy to see $a=b\neq 0$ which we insert into one of the equations to get $a(P/4-2a)=a^2$ with solution $a=P/12$ which gives $b=P/12$ and $c=P/4-2a=P/4-P/6=P/12$, i.e. all sides are of equal length, a cube.

Mikael Helin
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