# 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!

• Have you tried to test this for rectangular box of say, $8$ by $10$? Did it work? Oct 9, 2020 at 7:05
• @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 $8$. Oct 9, 2020 at 8:50
• 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. Oct 9, 2020 at 9:31
• 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? Oct 9, 2020 at 9:57
• @MathLover & Cosmo5: Thanks for your nice feedback. Very appreciated. Feb 21, 2021 at 15:49

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 $$\,0. The volume of the box is \begin{align} V(x)\; & =\; x(B-2x)(L-2x) \;=\; 4x\left(x-\frac B2\right)\left(x-\frac L2\right) \\[1.5ex] & =\; 4x^3 -2(B+L)\,x^2 +BL\,x\tag{1} \end{align} This cubic function is strictly positive if $$\,0 or $$\frac L2, and if $$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.

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