# Maximalization of a cubic puzzle

What is the maximal volume of a post package of length $L$, width $W$ and height $H$, subject to the following restrictions:

• $L+W+H \leq 90$
• $L \leq 60$, $W \leq 60$, $H \leq 60$

Intuitively I would say $30^3$, but how do I find the solution mathematically?

Taking the partial derivatives: $\ V_L=WH=V_W=LH=V_H=LW=0 =>L=W=H$

And therefore the maximum volume is: $V=(90/3)^3=30^3$. However, here I have assumed that the maximum volume occurs for $L+W+H=90$,

How do I prove this?

I have also not incorporated the $L \leq 60$, $W \leq 60$, $H \leq 60$ restrictions.

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Use the Arithmetic Mean Geometric Mean Inequality (AM-GM). If $x_1,x_2,\dots x_n$ are positive, then $$\frac{x_1+x_2+\cdots+x_n}{n}\ge (x_1x_2\cdots x_n)^{1/n},$$ with equality if and only if all the $x_i$ are equal.

For your problem, use the case $n=3$. Because $L+W+H \le 90$, by AM-GM we have $$(LWH)^{1/3}\le \frac{90}{3}=30,$$ with equality iff $L=W=H$. The given additional individual constraints on $L$, $W$, and $H$ make no difference, since they do not interfere with setting $L=W=H=30$.

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Thanks André Nicolas! It seems that the AM-GM states exactly what my intuition was telling me: "Thus the AM–GM inequality states that only the n-cube has the smallest sum of lengths of edges connected to each vertex amongst all n-dimensional boxes with the same volume." –  Andy Nov 24 '12 at 18:16