Spongy Cuboid

Let l,b and h be the length, breadth and height of the cuboid respectively.

From the given information,

$2(l + b) + lb = 16$ ... (1)

$2(b + h) + bh = 24$ ... (2)

$2(h + l) + hl = 31$ ... (3)

Solve (1) and (2) to get $(l –- b) (2 + h) = 7$ ...(4)

Solve (2) and (4) to get $4l - 5b = 2$

Solve (1) and last equation to get $b=2 , l=3$ and $c=5$

So the volume of the cuboid $=lbh = \boxed{30}$

Let the lengths of the distinct sides of the cuboid be $x,y,z$ .

We see that the information given to us is the same as the following system of equations:

$\left\{\begin{array}{l}xy+2x+2y=16\\ yz+2y+2z=24\\ zx+2z+2x=31\end{array}\right.$

Adding 4 to each of the equations:

$\left\{\begin{array}{l}xy+2x+2y+4=20\\ yz+2y+2z+4=28\\ zx+2z+2x+4=35\end{array}\right.$

Factoring by Simon's Favorite Factoring Trick

$\left\{\begin{array}{l}(x+2)(y+2)=20\\ (y+2)(z+2)=28\\ (z+2)(x+2)=35\end{array}\right.$

Multiplying all the equations together and square rooting gives $(x+2)(y+2)(z+2)=140$

Now we divide this by each of the equations to get:

$\left\{\begin{array}{l}x+2=5\\ y+2=4\\ z+2=7\end{array}\right.$ which means that $(x,y,z)=(3,2,5)$

Thus the volume of the cuboid is $3\times 2\times 5=\boxed{30}$

$xy + 2(x+y) = 16$

yz + 2(z+y) = 24

$zx + 2(z+ x) = 31$

Adding them,

$xy + yz + zx + 4(x+y+z) = 71$

Now,

$71 \geq 3{xyz}^{\frac{2}{3}} + 12{xyz}^{\frac{1}{3}}$

Let ${xyz}^{\frac{1}{3}} = a$

$0 \geq 3{a}^{2} + 12a -71$

Therefore, $a = \frac{-12 + \sqrt{144+852}}{6}$

a ~ 3.2

$xyz = {a}^{3} = 32.768$

Hence, maximum value of xyz is around 32.768

Therefore, xyz must lie between 28 and 35