← Differentiation methods

9 Optimisation

Method for an optimisation problem

  • Write the quantity to be optimised (area, volume, cost, time…) as a formula in TWO variables.
  • Use the given constraint to eliminate one variable, leaving a function of a SINGLE variable.
  • Differentiate, set the derivative to zero, and solve.
  • Confirm maximum or minimum with the second-derivative test or a sign table (see "Tangents, normals & curve features").
  • Check the domain — reject a negative length, an area outside the given range, or any solution the physical setup rules out.
  • Answer the actual question asked (a length, a cost, a time), with correct units — not just the value of the variable you solved for.
✕

A negative length is not a valid answer, even if the algebra gives one

Differentiating and solving can return more than one critical value, and it is common for one to be negative or outside the problem’s domain. Reject it before reporting your final answer — a negative side length or a negative time makes no physical sense, however clean the algebra looks. Worked example Maximise area A farmer has 40 m of fence for a rectangular pen against a wall (only 3 sides fenced). Maximise the area. Let width = x (two sides) and length = y (one side): 2x + y = 40 ⟹ y = 40 − 2x. Area A = xy = x(40 − 2x) = 40x − 2x². dA/dx = 40 − 4x = 0 ⟹ x = 10. (d²A/dx² = −4 < 0 ⟹ maximum.) y = 40 − 20 = 20. A = 10 × 20 = 200. ✓ Maximum area 200 m² (10 m × 20 m). Worked example Minimise cost A rectangular paddock of area 600 m² is to be fenced on all four sides. Fencing costs
2 per metre for the two sides of length x, and $8 per metre for the two sides of length y. Find the dimensions that minimise the total cost. Area: xy = 600, so y = 600/x. Cost: C = 2(12)x + 2(8)y = 24x + 16y = 24x + 16(600/x) = 24x + 9600/x. dC/dx = 24 − 9600/x². Setting dC/dx = 0: x² = 9600/24 = 400, so x = 20 (rejecting the negative root — a length cannot be negative). d²C/dx² = 19200/x³ > 0 for x > 0, confirming a minimum. y = 600/20 = 30. C = 24(20) + 9600/20 = 480 + 480 = 960. ✓ x = 20 m, y = 30 m, minimum total cost $960.

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