Finding the maximum without differentiation 


I have this problem that needs to be solved as if I was a GCSE student. 
After cutting a square of length $x$ from each corner, the volume of the open box would be $V= x(10-2x)(10-2x)$. We want to find the value of $x$ so that this volume is maximised. The ordinary solution would be to differentiate the cubic expression and find that the maximum point is at $x=\frac{5}{3}$. However, only knowledge up to KS4 (UK Year 11) may be used to solve it. Would the only appropriate approach be to use an iterative method (trial and error) to obtain an approximate value?

Note that by AM-GM inequality,
$$4V=4x(10−2x)(10−2x)\leq \left(\frac{4x+(10−2x)+(10−2x)}{3}\right)^3=\left(\frac{20}{3}\right)^3.$$
Equality holds, and the maximum value of $V$ is attained, when $4x=10-2x$, i.e. for $x=5/3$.
P.S. Actually what you need is just a particular case of AM-GM inequality: for $a,b\geq 0$,
$$ab^2\leq \left(\frac{a+2b}{3}\right)^3 \tag{1}$$
that is, after some standard calculations, equivalent to
$$(a+8b)(a-b)^2\geq 0$$
which holds. Equality is satisfied when $a=b$. Now apply $(1)$ with $a=4x$ and $b=10-2x$.
