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!

  • $\begingroup$ Have you tried to test this for rectangular box of say, $8$ by $10$? Did it work? $\endgroup$
    – Math Lover
    Oct 9, 2020 at 7:05
  • $\begingroup$ @MathLover, I was trying to compare general results using $L,B$ from calculus and this, but it quickly gets very messy due to square roots. I'm not sure about $10$ by $8$. $\endgroup$
    – cosmo5
    Oct 9, 2020 at 8:50
  • $\begingroup$ I asked because I was curious too :) I do not see anything wrong with your working. I don't think there is a straight forward way to do this using AM-GM for a rectangle. Taking a few examples and trying may help to see if it is working. $\endgroup$
    – Math Lover
    Oct 9, 2020 at 9:31
  • $\begingroup$ I'm not getting same $\lambda$ from inequality approach and calculus approach for $10$ by $8$. Maybe the assumption of optimum proportion $1:1/r:1/(4+2\lambda)$ is incorrect. But why? $\endgroup$
    – cosmo5
    Oct 9, 2020 at 9:57
  • 1
    $\begingroup$ @MathLover & Cosmo5: Thanks for your nice feedback. Very appreciated. $\endgroup$
    – Hanno
    Feb 21, 2021 at 15:49

1 Answer 1


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.

  • $\begingroup$ Very nice. Thank you for this answer. Yay to function/polynomial analysis! $\endgroup$
    – cosmo5
    Feb 20, 2021 at 8:07
  • $\begingroup$ I added a rewrite for $x_e$ in case it generates any insights. $\endgroup$
    – cosmo5
    Feb 20, 2021 at 8:39
  • $\begingroup$ Nice answer! Thanks for drawing my attention to it. Had forgotten all about it :) $\endgroup$
    – Math Lover
    Feb 20, 2021 at 14:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.