# Rectangular cuboid with volume equal to surface area?

If a rectangular cuboid's shortest edge is 3cm, and all the edge lengths are integers, how do you work out the longest edge if the volume of the cuboid is equal to the surface area?

• You cannot compare area and volume. It is like comparing speed to acceleration. Area has units $[L^2]$, and volume $[L^3]$, where $L$ is an unit of length. If you ignore units, you end up with meaningless garbage, because numbers themselves have no physical meaning. Basic Dimensional analysis is something everyone dealing with quantities should do automatically, to avoid e.g. crashing our very expensive interplanetary probes (kilometers versus miles). Sep 19, 2016 at 20:04
• You could fix the question by rewording it slightly, perhaps "the volume of the cuboid measured in cubic centimeters ($cm^3$) is equal to the surface area measured in square centimeters ($cm^2$)". This might look like nitpicking, but fact is, the units matter more than the numbers do. Just try to solve this problem using e.g. pints or milliliters for the volume, centimeters for the lengths, and square inches for the area. Sep 19, 2016 at 20:07
• I mean that the numerical value of the volume and surface area is the same. I know that the edges have lengths 3cm, 7cm and 42cm which gives a volume of 882cm cubed and a surface area of 882cm squared, but I need to find an algebraic way to work out the unknown lengths. Sep 19, 2016 at 20:07
• I think I now understand what the actual problem at hand is. For example, if the side lengths are $3 cm$, $8 cm$, and $24 cm$, then the volume is $576 cm^3$ and surface area $576 cm^2$. However, you are looking for the solution with all side lengths integers, and one of the side lengths is largest possible (integer). Sep 19, 2016 at 20:28
• Yes that's correct. I need to find the largest possible side length where the numerical value of the surface area and volume is the same. I know that the answer is 42cm, but I have no idea how you can work this out using algebra. Sep 19, 2016 at 20:42

We have a cuboid with side lengths $x$, $y$, and $z$. Its volume is $$V = x y z$$ and area is $$A = 2 x y + 2 x z + 2 y z$$

When the side lengths are measured in centimeters, we are looking for integer side lengths for which $$\frac{V}{1 \; cm^3} = \frac{A}{1 \; cm^2}$$ and one of the lengths is the largest possible (integer). We are given $x = 3 cm$.

From now on, for simplicity, we can drop the units. Just remember you cannot do unit conversions; we are assuming the above units.

(When you aren't sure if the method you are trying to apply to solve the problem is sane, you can always apply quick dimensional analysis to see if the dimensions match. If they do not, you are comparing apples and oranges, and most likely the method does not apply -- or more likely, you perhaps remember some details wrong. It is useful to know unit conversions, so you don't get tripped up if you have $[kg \, m/s^2]$ on one side, and $[N]$ on the other: they are equivalent. Some units, like specific impulse used to characterize rockets, may have weird units: if propellant weight is used, specific impulse is typically measured in seconds. In dimensional analysis, you work through the equations, but ignoring the numbers (unless you suspect you might be dividing by zero, or by infinity), and only look at the units. So, it's quite simple, and very powerful. It's saved my bacon a lot of times; I'm often wrong, but I sometimes catch myself, when checking my thinking using tools like dimensional analysis.)

The equation (for area and volume being equivalent) can now be written as $$x y z = 2 ( x y + x z + y z )$$ and substituting $x = 3$, $$3 y z = 6 y + 6 z + 2 y z$$ Subtracting $2 y z$ from both sides we get $$y z = 6 (y + z)$$ which is equivalent to $$y z - 6 y = 6 z$$ and $$y (z - 6) = 6 z$$ Dividing both sides by $(z - 6)$ we get $$y = \frac{6 z}{z - 6}$$

Note that due to symmetry, we also have $$z = \frac{6 y}{y - 6}$$ You can work out the previous three or four steps to verify it if you wish.

Because of the symmetry regarding $y$ and $z$, without loss of generality, we can declare we want $$z \ge y$$

When only integer solutions are asked for, it usually suffices to look at the general behaviour (derivatives, long-scale changes), and a small number of potential candidate answers near the interesting points indicated by the general behaviour (discontinuities, extrema i.e. where the derivative is zero, and so on). Occasionally the solutions are more complicated -- say, distances to 2D lattice points --, in which case additional work is needed.

If $y = 6$, then $z$ is undefined due to division by zero. If $y = 7$, then $z = 6 \cdot 7 / (7 - 6) = 42$.

For all $y \gt 7$, $z \lt 42$. One good way to verify this is look at the derivative of $6 y / (y - 6)$: $$\frac{d z}{d y} = \frac{6}{y - 6} - \frac{6 y}{(y - 6)^2} = -\frac{6^2}{(y - 6)^2}$$ Because the derivative is negative for all $y \gt 6$, it means $z$ is monotonically decreasing for $y \gt 6$: the values of $z$ become smaller and smaller as $y$ grows.

Because the largest $z$ is reached when $y = 7$ (then $z = 42$), and $x = 3$ was given, the solution is $$\begin{cases} x = 3 \; cm \\ y = 7 \; cm \\ z = 42 \; cm \end{cases}$$