All subjects · Mathematics · WMA12 (P2)

Optimisation problems

Differentiation finds the biggest or smallest value of a quantity, such as the largest area for a fixed length of fence.

Method

1. Write the quantity to optimise in terms of one variable. Use the constraint (fixed perimeter, volume, ...) to remove the other variable.

2. Differentiate and set the derivative to 0.

3. Use the second derivative to show it is a maximum or minimum.

4. Answer the question, with units.

Example of a constraint

A rectangle with perimeter 40 has sides x and 20 - x, so its area is A = x(20 - x) = 20x - x2.

Reject silly values

Lengths must be positive. If a solution gives a zero or negative length, reject it.

Worked example

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

  1. A = x(20 - x) = 20x - x2
  2. dA/dx = 20 - 2x = 0, so x = 10
  3. d2A/dx2 = -2 < 0, so a maximum
  4. A = 10 x 10

Answer: 100 cm2

Key idea: Use the constraint to get one variable, set the derivative to 0, and confirm with the second derivative.

Check yourself

  1. Two positive numbers add up to 30. Find the largest possible value of their product.
    Show answer

    225. P = x(30 - x). dP/dx = 30 - 2x = 0 gives x = 15, so P = 225.

  2. Squares of side x cm are cut from the corners of a 12 cm by 12 cm sheet, and the sides are folded up to make an open box with V = x(12 - 2x)2. Find the value of x that gives the maximum volume.
    Show answer

    2 cm. dV/dx = (12 - 2x)2 - 4x(12 - 2x) = (12 - 2x)(12 - 6x). x = 6 gives zero volume, so x = 2.

  3. For the same box, V = x(12 - 2x)2. Find the maximum volume.
    Show answer

    128 cm3. V = 2 x 82 = 128 cm3.

  4. A farmer has 60 m of fencing to make three sides of a rectangle against a wall. Find the maximum area.
    Show answer

    450 m2. With width x, the length is 60 - 2x and A = 60x - 2x2. dA/dx = 60 - 4x = 0 gives x = 15, so A = 15 x 30 = 450 m2.

  5. The cost of a process is C = 2x + 50/x, for x > 0. Find the minimum cost.
    Show answer

    20. dC/dx = 2 - 50/x2 = 0 gives x2 = 25, so x = 5 and C = 10 + 10 = 20.

  6. How do you show that a stationary value is a maximum?
    • Show that the second derivative is negative there
    • Show that the second derivative is positive there
    • Show that the second derivative is zero there
    • Show that the first derivative is positive there
    Show answer

    Show that the second derivative is negative there. A negative second derivative means the gradient is falling: a maximum.

Practise with instant feedback

Sources

Last updated 7 November 2026. Spotted a mistake? Tell us.