An open‑top rectangular box is to be made from a sheet of material having total surface area $1500{\text{cm}}^2$. Find the dimensions (length $l$, width $w$, height $h$) of the box that give the maximum possible volume. Give the dimensions correct to two decimal places.

Step-by-step solution

1. For an open top, surface area $S$ consists of the base and four side walls: $S=lw+2lh+2wh=1500$.
2. We want to maximise volume $V=lwh$ subject to the surface‑area constraint.
3. Using symmetry, the maximum occurs when the base is a square, i.e., set $l=w=x$. (Justification: for fixed $lw$, the product $lh\cdot wh$ is maximised when $l=w$.)
4. Replace $l$ and $w$ with $x$: the constraint becomes $x^2+4xh=1500$.
5. Express $h$ in terms of $x$: $h=\frac{1500-x^2}{4x}$.
6. Write volume in terms of $x$ only: $V=x^{2}h=x^2\left(\frac{1500-x^2}{4x}\right)=\frac{x(1500-x^2)}{4}=\frac{1500x-x^3}{4}$.
7. Differentiate $V$ with respect to $x$: $\frac{dV}{dx}=\frac{1500-3x^2}{4}$.
8. Set derivative to zero for a maximum: $1500-3x^2=0\Rightarrow x^2=500\Rightarrow x=\sqrt{500}=10\sqrt{5}\approx 22.36\text{cm}$.
9. Find $h$ using the expression for $h$: $h=\frac{1500-x^2}{4x}=\frac{1500-500}{4\times 22.36}=\frac{1000}{89.44}\approx 11.18\text{cm}$.
10. Thus the optimal dimensions are $l=w\approx 22.36\text{cm}$ and $h\approx 11.18\text{cm}$.
11. Check that the second derivative $\frac{d^{2}V}{d{x}^2}=-\frac{3x}{2}<0$ at $x=22.36$, confirming a maximum.

Concept tested: Maximum volume with a surface‑area constraint

Published 15 September 2026 · How this solution was prepared and checked