Maxima and minima in practical problems

Calculus finds the best size for a box or a can. The skill is writing the quantity in terms of one variable, then justifying that you have a maximum or minimum.

Practise this lesson free (6 questions)

The method

1. Write the quantity to be made largest or smallest.
2. Use the other information to write it in terms of one variable.
3. Differentiate and solve = 0.
4. Justify with the second derivative: < 0 maximum, > 0 minimum.
5. Answer the question asked (often the maximum value, not x).

An open box

Squares of side x are cut from the corners of a 12 cm by 12 cm sheet and the sides folded up.
V = x(12 − 2x)2
dVdx = (12 − 2x)2 − 4x(12 − 2x) = (12 − 2x)(12 − 6x) = 0
x = 6 gives no box, so x = 2. d2Vdx2 = -48 < 0, so V = 128 cm3 is a maximum.

A closed can

A closed cylinder holds 250 cm3. πr2h = 250, so h = 250πr2.
A = 2πr2 + 2πrh = 2πr2 + 500r
dAdr = 4πr − 500r2 = 0, so r3 = 125π and r = 3.41 cm.
d2Adr2 = 4π + 1000r3 > 0, so this is a minimum: A = 220 cm2.
Worked example

A rectangle has perimeter 40 cm. Find the maximum area.

  1. Width x, length 20 − x, so A = x(20 − x) = 20x − x2
  2. dAdx = 20 − 2x = 0, so x = 10
  3. d2Adx2 = −2 < 0, a maximum

Answer: A = 100 cm2

Key idea

Get the quantity in terms of one variable, differentiate, solve = 0, justify with the second derivative, then answer the question.

Key words
Second derivative
d^{2}y/dx^{2}, the derivative of dy/dx; used to decide the nature of stationary points.
Derivative
The gradient function dy/dx: the rate of change of y with respect to x.

All International GCSE Further Pure Mathematics key terms and definitions →

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Extra practice

2 more questions

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  1. Squares of side x cm are cut from the corners of an 18 cm square sheet to make an open box. Find the value of x that gives the maximum volume.

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    Answer: 3

    V = x(18 − 2x)2. dVdx = (18 − 2x)(18 − 6x) = 0 gives x = 3 (x = 9 makes no box).

  2. Squares of side x cm are cut from the corners of an 18 cm square sheet to make an open box. Find the maximum volume of the box.

    Show the answer

    Answer: 432 cm3

    The maximum is at x = 3, so V = 3 × 122 = 432 cm3.

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Sources and references

Checked by the GCSE-ready editorial team, 17 October 2026. Last updated 2 October 2026. Spotted a mistake? Tell us and we will fix it.

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