06 Maximum and minimum applications
Method: write the quantity to be optimised as a function of one variable using the constraint, state the admissible interval, differentiate, set f′ = 0, justify the nature, then answer the question in words with units.
- Fencing: a rectangular field of area 1,200 m² is to be fenced. Perimeter P = 2(x + 1200/x) ⟹ P′ = 2 − 2400/x² = 0 ⟹ x = √1200 ≈ 34.6 m, P minimum ≈ 138.6 m.
- Open box: from a 20 cm × 20 cm square, cut squares of side x and fold. V = x(20 − 2x)² ⟹ V′ = (20 − 2x)(20 − 6x) ⟹ x = 10/3 cm gives the greatest volume ≈ 592.6 cm³.
- Closest point: the point on y² = 4x nearest (3, 0): minimise D² = (x − 3)² + y² = (x − 3)² + 4x ⟹ d/dx = 2x − 2 = 0 ⟹ x = 1 ⟹ points (1, ±2), distance 2.
- Revenue: p = 100 − 0.5x, R = px = 100x − 0.5x² ⟹ R′ = 100 − x = 0 ⟹ x = 100 units, price EGP 50, R = EGP 5,000.