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Cuboid Challenge Poster

Age 11 to 14
Challenge Level

 

The largest volume you can make is when $x = 3\frac13$,
which gives a volume of $592\frac{16}{27}$ cm$^2$.

 

Why?

For any value of $x$, the other sides are each $20-2x$ and then the volume is $x\times(20-2x)\times(20-2x)$

Trying out values of $x$, a spreadsheet program like Excel is useful

$x$ $20-2x$ volume   $x$ $20-2x$ volume   $x$ $20-2x$ volume
2 16 512   2.5 15 562.5   3.1 13.8 590.36
3 14 588   3 14 588   3.2 13.6 591.87
4 12 576   3.5 13 591.5   3.3 13.4 592.55
5 10 500   3.6 12.8 589.82   3.4 13.2 592.42

You can keep going to get really close to 3.333333...

$x$ $20-2x$ volume   $x$ $20-2x$ volume
3.31 13.38 592.571   3.331 13.338 592.59237
3.32 13.36 592.585   3.332 13.336 592.59252
3.33 13.34 592.592   3.333 13.334 592.59259
3.34 13.32 592.591   3.334 13.332 592.59257

 

Using a cubic equation, the volume is $x(20-2x)^2$

which expands to $4x^3 - 80x^2 + 400x$

Differentiate to find the stationary points:

$\frac{\text{d}V}{\text{d}x} = 12x^2 - 160x + 400$

$\begin{align} 12x^2-160x+400 &= 0\\
\Rightarrow 3x^2 - 40x + 100 &= 0\\
\Rightarrow 3x^2 - 30x -10x + 100 &= 0\\
\Rightarrow 3x(x-10) - 10(x-10)&=0\\
\Rightarrow (3x-10)(x-10)&=0\\
\Rightarrow 3x-10 = 0 \text{ or } x-10 = 0\end{align}$

$x=10$ gives the minimum area of $0$, so $3x=10\Rightarrow x=\frac{10}3$ gives the maximum area.