Maximum capacity of box without using Calculus? Lets use a square cardboard paper (side $s$) to make an open box. Remove four small squares from each corner and fold to make a rectangular box of height $x$.
To compute maximum possible volume, we can use AM-GM :
$$ \sqrt[3]{(s-2x)(s-2x)\color{blue}{4}x} \le \dfrac{(s-2x)+(s-2x)+\color{blue}{4}x}{3} $$
$$ \Rightarrow V \le \dfrac{2}{27}s^3$$
Note : This matches the result obtained from calculus.
Next take a rectangular paper ($L \times B$) where $L > B$.
But AM-GM won't apply directly this time, since $L-2x > B-2x$.
This can be easily done by calculus. But I want to know if a valid solution using inequalities or other pre-Calculus methods exists?
I tried :
Let $(L-2x)=\lambda (B-2x)$ for maximum capacity. Then AM-GM,
$$ \sqrt[3]{(L-2x)\lambda(B-2x)(2+2\lambda)x} \le \dfrac{(L-2x)+\lambda(B-2x)+(2+2\lambda)x}{3} $$
$$ \Rightarrow 2\lambda(1+\lambda)V \le \frac{1}{27}(L+\lambda B)^3$$
where $$ (L-2x)=\lambda(B-2x)=(2+2\lambda)x $$
gives a quadratic in $\lambda$ : $$\dfrac{\lambda (\lambda+2)}{2\lambda+1} = \dfrac{L}{B} = r$$
One obtains
$$ V \le  \dfrac{(r+\lambda)^3}{\lambda (\lambda+1)} \dfrac{B^3}{54} $$

Is this correct? If it's incorrect, can this solution be improved? Or, does a different solution using pre-Calculus methods exist?

Thank you for your time!
 A: How to determine the height $x$ of the box maximising the box volume $V$ by a method which requires neither calculus nor inequality techniques.
Length $L$ and breadth $B$ are given and it is assumed that $\,0<B\leqslant L$. The volume of the box is
$$\begin{align}
V(x)\; & =\; x(B-2x)(L-2x)
 \;=\; 4x\left(x-\frac B2\right)\left(x-\frac L2\right) \\[1.5ex]
& =\; 4x^3 -2(B+L)\,x^2 +BL\,x\tag{1}
\end{align}$$
This cubic function is strictly positive if $\,0<x<\frac B2$ or $\frac L2<x$,
and if $x\leqslant 0$ or $\,\frac B2\leqslant x\leqslant\frac L2$
then $V(x)\leqslant 0\,$.
Note that $2(B+L)=P\,$ is the perimeter and $BL=A$ is the area of the given rectangular paper.
Now the idea is to shift the function $V$ parallel to the $y$-axis such that one extremum becomes a second order zero, i.e., the function graph touches the $x$-axis without changing sign. This preserves the $x$-coordinates of the extrema whereas the number of parameters needed to characterise the shifted function is reduced by one.
Let $x_e$ be the $x$-coordinate of one extremum, let $V_e=V(x_e)$, and let $u$ be an unknown. Then
$$\begin{align}V_\text{shifted}(x) \,=\, V(x)-V_e\: & \stackrel{!}{=}\:
4(x-u)(x-x_e)^2 \\[1ex]
& =\: 4x^3 -(4u+8x_e)\,x^2 +\big(8ux_e +4x_e^2\big)\,x - 4ux_e^2\tag{2}
\end{align}$$
Compare $(2)$ and $(1)$, then uniqueness of the coefficients implies
$\,4u=\dfrac{V_e}{x_e^2},$ and, exploiting this identity for $u$ straight away, also
$$\begin{align}2\,\frac{V_e}{x_e} +4x_e^2 & \;=\; BL\tag{3}\\[1.5ex]
\frac{V_e}{x_e^2} +8x_e & \;=\; 2(B+L)\,.\tag{4}\end{align}$$
Performing $\,(4)\cdot2x_e-(3)\,$ gives
$$x_e^2 -\frac{B+L}{3}x_e +\frac{BL}{12} \;=\; 0\tag{5}$$
with the solutions
$$x_e\;=\;\frac{B+L\mp\sqrt{BL+(L-B)^2}}{6}\,.$$
Clearly the smaller value belongs to the seeked maximum.
The larger value is the local minimum of $V\,$ between $\,\frac B2$ and $\,\frac L2$.
Cross-checking the equality case $L=B=s\,$ yields $\,x_e=\frac16s$, in agreement with the OP where this value corresponds to equality in AM-GM.
Rewriting the result in terms of $P=2(L+B)$ and $A=LB$ reads
$$x_e=\frac{P\mp \sqrt{P^2-12A}}{12}\,.$$

Finally, $V_e$ is made a bit more explicit: Performing $\,(3)\cdot 2-(4)\cdot x_e\,$ gives
$$V_e \;=\;\frac13\big[2BL\,x_e -2(B+L)\,x_e^2\big] 
 \;=\; \frac1{18}BL(B+L) -\frac29\big(BL+(L-B)^2\big)\,x_e$$
where $(5)$ has been used to replace $\,x_e^2\,$ by $\,\frac{B+L}{3}x_e -\frac{BL}{12}\,$.
Cross-checking the equality case $L=B=s\,$ gives $\,V_e=\dfrac2{27}s^3\,$ as desired.
