← 2022 Paper 2

UPSC 2022 Maths Optional Paper 2 Q3b — Step-by-Step Solution

20 marks · Section A

Maxima and minima of multi-variable functions (analytic criteria) · Real Analysis · asked 5× in 13 yrs · Read the full method →

Question

Find max/min of $\dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}+\dfrac{z^2}{c^4}$ subject to $lx+my+nz=0$ and $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1$. Interpret geometrically.

Technique

Two-constraint Lagrange multipliers; at critical points, $f=\mu$ (Lagrange multiplier); $\mu$ satisfies an equation derived from the plane constraint.

Solution

Setup. Two constraints (plane + ellipsoid surface), so two multipliers. Maximize/minimize $f$ subject to $g_1=lx+my+nz=0$ and $g_2=x^2/a^2+y^2/b^2+z^2/c^2-1=0$.

Step 1 — Lagrangian

$L=f-\lambda g_1-\mu g_2$.

$\partial L/\partial x=2x/a^4-\lambda l-2\mu x/a^2=0\Rightarrow 2x(1/a^4-\mu/a^2)=\lambda l\Rightarrow x=\dfrac{\lambda l a^4}{2(1-\mu a^2)}$.

Similarly $y=\dfrac{\lambda m b^4}{2(1-\mu b^2)}$, $z=\dfrac{\lambda n c^4}{2(1-\mu c^2)}$.

Step 2 — Substitute into plane constraint $g_1=0$

$lx+my+nz=0$:

Either $\lambda=0$ (trivial, gives $x=y=z=0$ which violates $g_2=0$ unless $a,b,c=0$), or the bracket is zero:

Step 3 — Substitute into ellipsoid constraint $g_2=0$

Alternatively, observe: at the critical point,

From Step 1, $x/a^2=\dfrac{\lambda l a^2}{2(1-\mu a^2)}$. So $\dfrac{x^2}{a^4}\cdot a^2=x^2/a^2=\dfrac{\lambda^2 l^2 a^4}{4(1-\mu a^2)^2}$.

Hmm, let me try a different approach: multiply the Lagrangian condition by $x$ (and similarly):

$x\partial L/\partial x=0$: $2x^2/a^4-\lambda lx-2\mu x^2/a^2=0$.

Summing over $x,y,z$ and using $lx+my+nz=0$: $2f-2\mu\cdot 1=0$ (where $f=\sum x^2/a^4$, and $\sum x^2/a^2=1$).

So $f=\mu$ at the critical point.

This means the extremum value of $f$ equals the Lagrange multiplier $\mu$.

Step 4 — Determine $\mu$ from $(\star)$

The equation $(\star)$ is a cubic in $1/\mu$ (after clearing denominators). The two roots correspond to the max and min values.

Actually, multiplying $(\star)$ by $\prod(1-\mu a^2)$ etc. gives a cubic equation:

The roots $\mu_1,\mu_2$ (and possibly $\mu_3$) are candidate extrema. Two real roots typically; the larger is max, smaller is min.

Standard result: $\mu_{\max}$ and $\mu_{\min}$ are the two roots of a quadratic obtained by simplifying $(\star)$.

Step 5 — Geometric interpretation

The plane $lx+my+nz=0$ passes through the origin and cuts the ellipsoid in an ellipse. The function $f=x^2/a^4+y^2/b^4+z^2/c^4$ measures (up to scaling) the inverse-square distance from the centre along the ellipsoid’s principal axes.

Geometric meaning: $1/\sqrt f$ is the length of the semi-axis of the planar section’s ellipse. The max and min of $f$ correspond to the shortest and longest semi-axes of the elliptic section.

So the answer can be stated as:

• Max value of $f$ = $1/(\text{shortest semi-axis})^2$ of the elliptic section.
• Min value of $f$ = $1/(\text{longest semi-axis})^2$.

Answer