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MathidiaMathematics Academy
Differentiation
Lesson 35 of 75
Review

Optimisation

Objective. Form and solve constrained optimisation models.

  1. 1. Learn
  2. 2. Worked example
  3. 3. Practice
  4. 4. Feedback
  5. 5. Continue

Learn: the key idea

Optimisation uses calculus to find the maximum or minimum value of a quantity subject to a constraint, by writing the quantity as a function of one variable, differentiating, and setting the derivative to zero. First express the objective in terms of a single variable using the given constraint to eliminate others, then solve f'(x) = 0 and justify using the second derivative or context whether it gives a maximum or minimum. For example, to maximise the area of a rectangle with perimeter 20, write A = x(10 − x) = 10x − x², giving dA/dx = 10 − 2x = 0, so x = 5 and A = 25. Always check the answer makes sense within the practical constraints of the problem.

Worked example

A farmer has 40 m of fencing to enclose a rectangular field against an existing wall, so fencing is only needed on three sides. Use optimisation to find the maximum enclosed area.

Diagram of a rectangle
  1. 1Let the two equal sides be x, so the side parallel to the wall is 40 − 2x; area A = x(40 − 2x) = 40x − 2x².
  2. 2Differentiate: dA/dx = 40 − 4x, set to 0 giving x = 10.
  3. 3Check it's a maximum: d²A/dx² = −4 < 0, confirming a maximum.
  4. 4Compute area: A = 10×(40 − 20) = 10×20 = 200 m².

Answer: 200 m²

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Common mistake. Students often optimise a function of two variables directly without first using the constraint to reduce it to one variable, making differentiation impossible. Always substitute the constraint in before differentiating.

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