# What dimensions would you need to make a glass cage with maximum volume if you have a piece of glass that is 14" by 72"?

Feb 24, 2015

If we are allowed to cut and attach to each other pieces of glass of any size, we have to make a cubical cage with the area of all sides combined equal to the area of a piece of glass given, that is $14$x$72$. It is rather strict assumption and it's not obvious from the problem whether it is allowed. Assume, it is.
Then here are the detailed proof and calculations.

Assuming the dimensions of a cage are $X$, $Y$ and $Z$, we can formulate a problem as follows:
Find $X$, $Y$ and $Z$ such that $V = X \cdot Y \cdot Z$ (volume of the cage) reaches its maximum while $S = 2 \cdot \left(X \cdot Y + Y \cdot X + Z \cdot X\right)$ (combined area of all 6 sides of a cage) is given. In our case $S = 14 \cdot 72$ square inches.

As we noted above, we assume that all glass can be used to make a cage without any difficulties of cutting and attaching pieces together.

As we know, arithmetic average of a group of numbers is greater than or equal to their geometric average with equality reached only if all numbers are equal to each other.
Therefore, we can state that
${\left[\left(X \cdot Y\right) \cdot \left(Y \cdot Z\right) \cdot \left(Z \cdot X\right)\right]}^{\frac{1}{3}} \le \frac{\left(X \cdot Y\right) + \left(Y \cdot Z\right) + \left(Z \cdot X\right)}{3}$

Using expressions for volume $V$ (unknown to be maximized) and area of all sides $S$ of a cage (known constant), we can represent the above inequality as
${\left({V}^{2}\right)}^{\frac{1}{3}} \le \frac{S}{6}$ or
$V \le \sqrt{{\left(\frac{S}{6}\right)}^{3}}$, where $S = 14 \cdot 72$,
with equality reached only if
$X \cdot Y = Y \cdot Z = Z \cdot X$, that is if
$X = Y = Z$, that is our cage must be a cube.

We have to find an edge of a cube whose side area equals to $14 \cdot 72$.
Solve the equation:
$6 \cdot {X}^{2} = 14 \cdot 72$ or ${X}^{2} = 14 \cdot 12 = 168$
Therefore, dimensions we are seeking are
$X = Y = Z = \sqrt{168}$

This dimension of a side, which is about $12.96$, obviously presents a practical challenge. It will be necessary to cut and attach different pieces of glass to make the cubical cage of this dimension, and it will look like a Tiffany lamp. But beauty is in the eyes of a beholder.
At least, mathematically, it will really be a cage with a maximum volume made from a given piece of glass.