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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-2/3d/solution

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Past exam of the mathematics course of the University of Cambridge / 2020 / ib / Paper 2 / 3D / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-29

Because the constraint determines

x = b^{2} - a^{2} y^{2} - z^{2},

the constrained function is

Φ (y, z) = yz (b^{2} - a^{2} y^{2} - z^{2}) .

Its stationary points satisfy

\frac{\partial Φ}{\partial y} = z (b^{2} - 3 a^{2} y^{2} - z^{2}) = 0, \frac{\partial Φ}{\partial z} = y (b^{2} - a^{2} y^{2} - 3 z^{2}) = 0.

When

yz = 0

, these equations give

(x, y, z) = (b^{2}, 0, 0), (0, \pm b / a, 0), (0, 0, \pm b) .

When

yz \neq = 0

, subtracting the two bracketed equations gives

z^{2} = a^{2} y^{2}

, and substitution then gives

x = \frac{b ^{2}}{2}, y = \pm \frac{b}{2 a}, z = \pm \frac{b}{2},

with the two signs chosen independently. These are exactly the constrained stationary points; equivalently they follow from the Lagrange multiplier equations.

Under the additional restriction

x \geq 0

, the constraint gives the compact elliptical disc

a^{2} y^{2} + z^{2} \leq b^{2}

. On its boundary

x = 0

, the objective is zero. At the four nonzero interior stationary points,

ϕ = x yz = \pm \frac{b ^{4}}{8 a},

with the sign determined by

yz

. Therefore

max ϕ = \frac{b ^{4}}{8 a}, min ϕ = - \frac{b ^{4}}{8 a} .

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