Calculate Interior Dimensions Using Cubic Feet I have been tasked to calculate the interior dimensions of a product given some non-traditional values. Given the following numerical information how would one go about calculating the interior dimensions of an object (let's say a food storage freezer box in this case): 
The exterior dimensions are provided as $41''$ height $\times 22''$ width $\times 31''$ depth and the cubic footage is stated as $7.3$ (I understand CF is a measurement of volume). 
Is there a way to work with this information to devise a mathematical equation to determine the approximate interior dimensions of this object?
 A: We will have to make some assumptions. First, let us assume that the exterior dimensions given are with the lid on. Second, let us assume that the side walls, the bottom, and the lid are all made of insulating material of the same thickness.
Let that thickness be $t$. Then the interior dimensions, in inches, are $41-2t$, $22-2t$, and $31-2t$. The interior volume, in cubic inches, is therefore 
$$(41-2t)(22-2t)(31-2t).$$
We are told that this interior volume is $7.3$ cubic feet. Since there are $12$ inches in a foot, the volume, in cubic inches, is $(7.3)(12^3)$. It follows that
$$(41-2t)(22-2t)(31-2t)=(7.3)(12^3).\tag{1}$$
Our problems are not over, for our equation is a cubic equation. There is a formula for solving cubics, but it is quite complicated, and in my opinion not useful for your problem. So we need to solve Equation (1) numerically.
There are many options. If you have a graphing calculator, or a graphing program, you can look at the function $f(t)=(41-2t)(22-2t)(31-2t)-(7.3)(12^3)$ and read off, approximately, where it is $0$.
Some calculators have a Solve button that will solve equations numerically, to good accuracy.  
The effectively free program Wolfram Alpha will also do it. 
There are many other possibilities. One of them, not as painful as you might think, is to calculate $(2t-41)(2t-22)(2t-31)$ for various values of $t$. You will fairly quickly pin down the $t$ that works to good accuracy. As a starting hint, the $t$ is bigger than $3$.
Once you have found the suitable $t$, subtract $2t$ from each of the exterior dimensions to find the corresponding interior dimensions.  
