A-Level Differentiation: worked solution

Question

For the right hexagonal pyramid with V²=(3/4)(a²x⁴ − x⁶), the volume is greatest when x²=(2/3)a². Find, in terms of a, the total surface area at that maximum-volume configuration. [4]

First identify what a is, by matching the given V² to the geometry. A regular hexagon of side x splits into six equilateral triangles of side x, so base area = 6 × (√3/4)x² = (3√3/2)x². For a right pyramid of height h, V = (1/3)(base area)(h) = (1/3)(3√3/2)x²h = (√3/2)x²h so V² = (3/4)x⁴h². Comparing with the given V² = (3/4)(a²x⁴ − x⁶) = (3/4)x⁴(a² − x²): h² = a² − x², so a is the slant (lateral) edge, since the hexagon's circumradius (centre to a vertex) is x and h² + x² = a². At the maximum-volume configuration, x² = (2/3)a², so h² = a² − (2/3)a² = (1/3)a². Base area = (3√3/2)x² = (3√3/2)(2/3)a² = √3 a². For the six triangular faces, each has base x and slant height l measured from the apex to the midpoint of a base edge. That midpoint sits at the apothem distance from the centre, and for a regular hexagon of side x the apothem is (√3/2)x. By Pythagoras in the vertical right triangle, l² = h² + ((√3/2)x)² = h² + (3/4)x² = (1/3)a² + (3/4)(2/3)a² = (1/3)a² + (1/2)a² = (5/6)a² l = a√(5/6). Also x = a√(2/3). Lateral area = 6 × (1/2) × x × l = 3xl = 3 × a√(2/3) × a√(5/6) = 3a²√(10/18) = 3a²√(5/9) = 3a² × (√5/3) = √5 a². Total surface area = base + lateral = √3 a² + √5 a² = (√5 + √3)a² units².