UPSC Maths 2025 Paper 2 Q3b — Solution

20 marks · Section A

Question

Show that the volume of the greatest rectangular parallelopiped that can be inscribed in the ellipsoid $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} = 1$ is $\dfrac{8abc}{3\sqrt{3}}$ .

Technique

Use Lagrange multipliers, exploiting the symmetry that a maximal inscribed box centred at the origin has vertices $(\pm x,\pm y,\pm z)$ .

Solution

By the central symmetry of the ellipsoid, the largest inscribed rectangular box is centred at the origin with edges parallel to the axes and one vertex $(x,y,z)$ in the first octant ( $x,y,z>0$ ) lying on the surface. Its edge lengths are $2x,2y,2z$ , so the volume is $V = (2x)(2y)(2z) = 8xyz,$ to be maximized subject to $g(x,y,z) = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} - 1 = 0.$

Step 1 — Lagrange conditions. Setting $\nabla V = \lambda\,\nabla g$ : $8yz = \lambda\,\frac{2x}{a^2},\qquad 8xz = \lambda\,\frac{2y}{b^2},\qquad 8xy = \lambda\,\frac{2z}{c^2}.$

Step 2 — Solve. Multiply the first equation by $x$ , the second by $y$ , the third by $z$ : $8xyz = \lambda\,\frac{2x^2}{a^2} = \lambda\,\frac{2y^2}{b^2} = \lambda\,\frac{2z^2}{c^2}.$ Since $8xyz\ne0$ at a maximum, $\lambda\ne0$ , so the three right-hand sides are equal: $\frac{x^2}{a^2} = \frac{y^2}{b^2} = \frac{z^2}{c^2}.$ Call this common value $t$ . Substituting into the constraint $g=0$ : $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 3t = 1 \;\Rightarrow\; t = \frac13.$ Hence $\frac{x^2}{a^2} = \frac13 \Rightarrow x = \frac{a}{\sqrt3},\qquad y = \frac{b}{\sqrt3},\qquad z = \frac{c}{\sqrt3}.$

Step 3 — Maximum volume. $V_{\max} = 8\cdot\frac{a}{\sqrt3}\cdot\frac{b}{\sqrt3}\cdot\frac{c}{\sqrt3} = \frac{8abc}{3\sqrt3}.$

Step 4 — This is a maximum. $V$ is continuous on the compact first-octant portion of the ellipsoid; it vanishes on the boundary (where any of $x,y,z\to0$ ) and is positive in the interior, so the unique interior critical point is the global maximum.

Rationalising, $\dfrac{8abc}{3\sqrt3} = \dfrac{8\sqrt3\,abc}{9}$ .

Answer

$\boxed{\;V_{\max} = \frac{8abc}{3\sqrt3} = \frac{8\sqrt3\,abc}{9},\quad\text{attained at } x=\frac{a}{\sqrt3},\ y=\frac{b}{\sqrt3},\ z=\frac{c}{\sqrt3}.\;}$