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!
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$\begingroup$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
$\endgroup$ 1 $\begingroup$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$$a=b=c.$$
$\endgroup$ 6 $\begingroup$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.
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