Real-Life Optimisation Problems

Read what is to be maximised or minimised; assign variables with units; write the constraint; express the objective in one variable; find derivative and critical value; test the result; interpret the answer in words with units.

A diagram may help for geometry-based word problems, but the essential requirement is a labelled variable model and constraint equation.

Finding x = 10 correctly in a rectangle problem but stopping there, even though the question asks for the maximum area or for both dimensions. Another realistic wrong answer is accepting a negative length or radius because the feasible domain was not checked.

Optimisation uses derivatives to find the maximum or minimum value of a quantity under given conditions or constraints.

Optimisation uses derivatives to find the maximum or minimum value of a quantity under given conditions or constraints. Optimisation method: form objective function Q; use constraint to express Q in one variable; solve Q'(x) = 0 within the feasible domain; verify maximum or minimum by second derivative test, sign test, or endpoint comparison where needed. For a rectangle with perimeter 20 cm, let length be x and breadth be 10 - x. Area A = x(10 - x) = 10x - x^2. A' = 10 - 2x, so x = 5. Since A'' = -2 < 0, area is maximum when x = 5 and breadth = 5; the rectangle is a square.

A real-life optimisation problem is first a modelling problem and only then a differentiation problem. The question usually gives a constraint, such as fixed perimeter, fixed volume, fixed surface area, or a relation between dimensions. The quantity to be maximised or minimised must be written as an objective function in one variable, using the constraint to eliminate the other variable. The feasible domain matters because lengths, radii, heights, areas, and costs cannot take arbitrary values. After solving Q'(x) = 0, students must verify whether the critical value gives a maximum or minimum by using Q''(x), sign change, or endpoint comparison, and then answer the practical question with correct units. Marks are often lost not in differentiation, but in choosing the wrong objective or forgetting to interpret the result. Optimisation method: form objective function Q; use constraint to express Q in one variable; solve Q'(x) = 0 within the feasible domain; verify maximum or minimum by second derivative test, sign test, or endpoint comparison where needed. A rectangular sheet has perimeter 40 cm. Find the rectangle of maximum area. Let length be x cm and breadth be y cm. Constraint: 2x + 2y = 40, so y = 20 - x. Area A = xy = x(20 - x) = 20x - x^2, where 0 < x < 20. Differentiate: A'(x) = 20 - 2x. Set A'(x) = 0: 20 - 2x = 0, so x = 10. A''(x) = -2 < 0, so the area is maximum. Then y = 20 - 10 = 10. The rectangle of maximum area is 10 cm by 10 cm. Usually appears as a long-answer or case-study problem involving maximum area, maximum volume, minimum cost, minimum surface area, or shortest distance. Finding x = 10 correctly in a rectangle problem but stopping there, even though the question asks for the maximum area or for both dimensions. Another realistic wrong answer is accepting a negative length or radius because the feasible domain was not checked.