2023 ACJC Promo Q9

[It is given that the volume of a pyramid is \(\frac{1}{3}\times \)base area\(\times \)height.]

A container in the shape of a right pyramid with vertical height \(H\)cm consists of a square base of side length \(x\)cm and four isosceles triangles, each with base \(x\)cm and perpendicular height \(h\)cm (see diagram). The pyramid has a fixed surface area \(A\)cm\(^{2}\).

1. Show that \(h=\frac{A-{{x}^{2}}}{2x}\).[1]
2. Hence show that \(36{{V}^{2}}={{A}^{2}}{{x}^{2}}-2A{{x}^{4}}\), where \(V\) is the volume of the pyramid.[3]
3. Find, in terms of \(A\), the value of \(x\) for which \(V\) is maximum, and show that this maximum value of \(V\) is \(\sqrt{\frac{{{A}^{3}}}{288}}\). You do not need to show that this value is a maximum.[3]