Minimise material in a thick lidless container

Original lidless tank geometryNative3D wireframe of open square-base tank; four1cm thickness labels show opposing walls and base; internal widthxcm; internal verticaledges dashed.x cm1 cm1 cm1 cm1 cm
Question diagram 1
Original lidless tank geometryNative3D wireframe of open square-base tank; four1cm thickness labels show opposing walls and base; internal widthxcm; internal verticaledges dashed.x cm1 cm1 cm1 cm1 cm

Given lidless container with internal side x and1cm wall/bottom thickness.

A manufacturer wants to design a lidless container with a square base and a capacity of 1000 cm31000\,\mathrm{cm}^3. Given that the uniform thickness of the container is 1 cm, using differentiation, find the internal length of the tank, xx cm, such that the volume of the material used is minimised. marks

[2009/MJC/Prelim/I/11]

Solution

Optimal internal side

Optimisation

A lidless container has a square base and capacity 1000 cm31000\ \mathrm{cm}^3; its walls and base are 1 cm thick. Using differentiation, find the internal length xx cm that minimises the volume of material.

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Hint

  1. Let the internal height be hh: x2h=1000x^2h=1000.

  2. The outer box has length x+2x+2 and height h+1h+1 (no lid, so one thickness at the bottom only).

  3. Material == outer volume −- inner volume; differentiate.

Answer · Optimal internal side

x=20003≈12.6x=\sqrt[3]{2000}\approx12.6 cm (3 s.f.).

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Same capacity. How much material?

Widen the square base and watch the height shrink. Find the balance that uses the least material, with no lid on top.

Open container cross-sectionOpen topInternal side: 7.00 cm

View fitted to the container. Four walls and one base; no lid.

Material volume (cm³)

Horizontal axis: internal side x (cm). The selected point follows the container.

Internal side x7.00 cm
Internal height
20.41 cm
h = capacity / x²
Capacity (fixed)
1000 cm³
Material used
734.06 cm³

178.25 cm³ above the minimum. Widening the base reduces the material used.