# For each face of a cuboid, the sum of perimeter and area is known. Find the volume.

I came across this problem in a national level examination.

On each face of a cuboid, the sum of its perimeter and its area is written. Among the six numbers written are $$16$$, $$24$$, and $$31$$.

Then the volume lies between ...

My workout...

From the problem,

$$ab+2(a+b)=16$$

$$bc+2(b+c)=24$$

$$ca+2(c+a)=31$$

I am sure some amount of manipulation is required after this step, but I have tried in vain.

Add $4$ to each equation and factorise \begin{eqnarray*} (a+2)(b+2)=20 = 5 \times 4 \\ (b+2)(c+2)=28 = 4 \times 7 \\ (c+2)(a+2)=35 = 7 \times 5 \\ \end{eqnarray*} which gives $\color{red}{a=3,b=2,c=5}$. So the volume is $\color{blue}{30}$.

• We can also show easily that this is the only solution by considering small variations in $a,b,c$. +1 Oct 21, 2017 at 14:13
• You can solve directly by multiplying the first two equations and dividing by the third: $$\frac{(a+2)(b+2)\cdot(b+2)(c+2)}{(c+2)(a+2)} = \frac{20\cdot 28}{35} \;\to\; (b+2)^2 = 16 \;\to\; b+2 = \pm 4 \;\to\; b = 2$$ (taking $b=-6$ to be extraneous).
– Blue
Oct 21, 2017 at 19:31

I much prefer the cleverness of @Donald's answer (which is probably the approach the exam writers hoped you'd see), but it's perhaps worth noting that you can attack this problem with a little algebraic brute force and a lot of perseverance.

You have three equations in three unknowns. Our goal is to eliminate two of the unknowns, leaving a single equation in, say, $a$. Notice that the first equation fairly readily allows us to express $b$ in terms of $a$; the third equation allow us to to do likewise for $c$:

\begin{align} ab + 2 a + 2 b = 16 \quad\to\quad b(a+2) = 16-2a \quad\to\quad b &= \frac{16-2a}{a+2} = \frac{2(8-a)}{a+2} \tag{1a} \\[6pt] c &= \frac{31-2a}{a+2} \tag{1b} \end{align}

By substitution, the second equation transforms to involve $a$ alone: $$\frac{2(8-a)}{a+2}\cdot\frac{31-2a}{a+2} + 2\left(\frac{16-2a}{a+2}+\frac{31-2a}{a+2}\right) = 24 \tag{2a}$$ $$\frac{(8-a)(31-2a)}{(a+2)^2}+ \frac{47-4a}{a+2} = 12 \tag{2b}$$ $$(8-a)(31-2a)+ (47-4a)(a+2) = 12 (a+2)^2 \tag{2c}$$

(It's around this point that you should start to suspect that there's a better way. Nevertheless, ...) We can expand everything and combine terms to get $$14 a^2 + 56 a - 294 = 0 \quad\to\quad 14 ( a^2 + 4 a - 21)= 0 \quad\to\quad 14 (a+7)(a-3) = 0 \tag{3}$$ Here, we solve to find that $a=3$ (discarding $a=-7$ as extraneous). Then $b=2$ and $c=5$ follow from $(1a)$ and $(1b)$ above. $\square$

This approach lacks ingenuity, but it's essentially mechanical. (You could even opt to use the Quadratic Formula in $(3)$ instead of thinking-through the factorization. Simplifying the final result is something of a chore, but it doesn't require any insight.) So, even if you don't see the clever approach, there's still a way to proceed ... and to succeed.

It was already done (and far better that what i did) but i isolated a b c in the next way:

a=(16-2b)/(b+2)

b= (24-2c)/(c+2)

c=(31-2a)/(a+2)

then by substitutions you will get the solution.