A-Level Differentiation: worked solution

Question

For the right hexagonal pyramid with V²=(3/4)(a²x⁴ − x⁶) and fixed edge a, find, in terms of a, the greatest volume achievable. (You need not verify it is a maximum.) [4]

V is greatest exactly when V² is greatest (V is a volume, so V > 0 and squaring preserves order), and V² is far easier to differentiate than V itself. So maximise W = V². W = (3/4)(a²x⁴ - x⁶), with a a fixed constant and x the variable. Differentiate with respect to x: dW/dx = (3/4)(4a²x³ - 6x⁵) For a stationary value, set dW/dx = 0: (3/4)(4a²x³ - 6x⁵) = 0 4a²x³ - 6x⁵ = 0 Factorise, taking out the common factor 2x³: 2x³(2a² - 3x²) = 0 So x³ = 0 or 2a² - 3x² = 0. x = 0 gives V = 0, the degenerate case with no pyramid, so take the other factor: 3x² = 2a² x² = (2/3)a² Substitute back into W = V². Write the powers of x in terms of x²: x⁴ = (x²)² = (2/3)²a⁴ = (4/9)a⁴ x⁶ = (x²)³ = (2/3)³a⁶ = (8/27)a⁶ V² = (3/4)( a² · (4/9)a⁴ - (8/27)a⁶ ) = (3/4)( (4/9)a⁶ - (8/27)a⁶ ) Put the bracket over 27: (4/9)a⁶ = (12/27)a⁶, so the bracket is (12a⁶ - 8a⁶)/27 = 4a⁶/27 V² = (3/4) × (4a⁶/27) = 3a⁶/27 = a⁶/9 Take the positive square root, since V is a volume: V = √(a⁶/9) = a³/3 So the greatest volume is V = a³/3 units³, attained when x² = (2/3)a².