Optimisation problems

Method: name the quantity to optimise, write it as a function of one variable using the given constraint, find its critical points, test them, and state the answer in the words of the question.

Worked example: poles 16 m and 22 m tall stand 20 m apart, and a stake is placed x m from the shorter one. The sum of squared wire lengths is S = 2x² − 40x + 1140, least at x = 10 m, the midpoint.

Worked example: a trapezium has three sides of 10 cm. If the longer parallel side is 10 + 2x, its area is A = (10 + x)√(100 − x²). A′ = 0 gives x = 5 with A″(5) = −30/√75 < 0, so the greatest area is 75√3 cm².

Worked example: from a 3 m × 8 m sheet, squares of side x are cut from the corners and the sides folded up. V = x(8 − 2x)(3 − 2x) = 4x³ − 22x² + 24x, V′ = 4(x − 3)(3x − 2). Only x = 2/3 fits the sheet, and it gives V = 200/27 m³.

Worked example: the point of y = x² nearest to (0, c), for c ≥ 1/2, is at height k = (2c − 1)/2, and the least distance is √(4c − 1)/2. Minimising the square of a distance is easier and gives the same point.

Worked example: a helicopter flies along y = x² + 7 and a soldier stands at (3, 7). Squared distance (x − 3)² + x⁴ has derivative 2(x − 1)(2x² + 2x + 3), and the quadratic has no real root, so x = 1 gives the least distance √5.

Worked example: the cylinder inscribed in a cone of radius r that has the greatest curved surface has radius r/2.

Always check the domain the problem allows. A root outside it, such as x = 3 for the sheet, is rejected.