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UPSC 2021 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 14 yrs · Read the full method →

Question

Find stationary values of $x^2+y^2+z^2$ subject to $ax^2+by^2+cz^2=1$ and $lx+my+nz=0$. Interpret geometrically.

Technique

Two-constraint Lagrange; "$r^2=\lambda$" via the standard scaling argument; stationary values satisfy a quadratic from the plane condition.

Solution

Setup. Two constraints — Lagrange multipliers with two multipliers $\lambda,\mu$.

$L=x^2+y^2+z^2-\lambda(ax^2+by^2+cz^2-1)-\mu(lx+my+nz)$.

Step 1 — Stationary conditions

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

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

Step 2 — Apply plane constraint $lx+my+nz=0$

$\dfrac{\mu}{2}\!\left[\dfrac{l^2}{1-\lambda a}+\dfrac{m^2}{1-\lambda b}+\dfrac{n^2}{1-\lambda c}\right]=0$.

Either $\mu=0$ (trivial, gives $x=y=z=0$, fails ellipsoid constraint), or

This is a quadratic in $\lambda$ (after clearing denominators). The two roots $\lambda_1,\lambda_2$ are the candidate stationary values.

Step 3 — Identify $\lambda=r^2$ at stationary points

Multiply each Lagrange equation by the variable and sum:

$\sum x\cdot 2x=\lambda\sum 2ax\cdot x+\mu\sum lx$:

$2\sum x^2=2\lambda\sum ax^2+\mu\sum lx$.

Use ellipsoid: $\sum ax^2=1$. Use plane: $\sum lx=0$.

$2r^2=2\lambda+0$, where $r^2=x^2+y^2+z^2$.

So $\lambda=r^2$ at stationary points.

Step 4 — Stationary values

Stationary value of $r^2=\lambda$, where $\lambda$ is a root of $(\star)$.

So the two stationary values $r_1^2,r_2^2$ are the two roots of (the quadratic from clearing denominators in) $(\star)$.

Answer