Stationary points and optimisation

Maths: Applications and Interpretation · Calculus. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

A stationary point is a point where the gradient of the curve is zero, so the tangent is horizontal. To find one, differentiate, set dy/dx = 0 and solve for x. Then put x back into the original equation to get y.

At a local maximum the gradient changes from positive to negative (up, then down). At a local minimum it changes from negative to positive. Test the sign of dy/dx just before and just after the point, or look at the graph on your calculator.

In optimisation you find the largest or smallest value of a quantity such as area, volume, cost or profit. Write the quantity as a function of one variable, then find its stationary point.

Rules to remember

Common mistake

Students stop at the x-value when the question asks for the maximum or minimum VALUE. Substitute x back into the original function (not into the derivative, which just gives 0) to get the area, volume or profit.

Worked example

An open box with a square base of side x cm must hold 32 cm3. Its surface area is S = x2 + 128/x. Find the minimum surface area.

  1. Write S = x2 + 128x−1, so dS/dx = 2x − 128x−2
  2. Set dS/dx = 0: 2x = 128/x2, so x3 = 64 and x = 4
  3. Check: at x = 2, dS/dx = 4 − 32 = −28 (negative); at x = 8, dS/dx = 16 − 2 = 14 (positive), so it is a minimum
  4. S = 42 + 128 ÷ 4 = 16 + 32 = 48

Answer: Minimum surface area = 48 cm2, when x = 4 cm.

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