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UPSC 2015 Maths Optional Paper 2 Q4b — Step-by-Step Solution

15 marks · Section A

Maxima and minima of single-variable functions · Real Analysis · asked 3× in 14 yrs · Read the full method →

Question

Find the absolute maximum and minimum values of the function $f(x,y)=x^2+3y^2-y$ over the region $x^2+2y^2\le 1$.

Technique

Interior critical point via gradient = 0; boundary via Lagrange multipliers; compare all candidate values to find absolute extrema.

Solution

Strategy. Continuous $f$ on a compact set (closed bounded ellipse interior) attains its extrema. Check (i) interior critical points, (ii) boundary $x^2+2y^2=1$ via Lagrange.

Step 1 — Interior critical points

$\nabla f=(2x,\;6y-1)=0\Rightarrow x=0,\;y=1/6$.

Check: $x^2+2y^2=0+2/36=1/18<1$ ✓ — inside the region.

$f(0,1/6)=0+3/36-1/6=1/12-2/12=-1/12$.

Step 2 — Boundary $x^2+2y^2=1$ via Lagrange

Maximize/minimize $f$ subject to $g=x^2+2y^2-1=0$.

$\nabla f=\lambda\nabla g$:

• $2x=\lambda\cdot 2x\Rightarrow x(1-\lambda)=0$, so $x=0$ or $\lambda=1$.
• $6y-1=\lambda\cdot 4y\Rightarrow (6-4\lambda)y=1$.

Case A: $x=0$. From constraint $2y^2=1\Rightarrow y=\pm 1/\sqrt 2$.

• $f(0,1/\sqrt 2)=0+3/2-1/\sqrt 2=1.5-0.707\approx 0.793$.
• $f(0,-1/\sqrt 2)=0+3/2+1/\sqrt 2=1.5+0.707\approx 2.207$.

Exact: $f(0,1/\sqrt 2)=3/2-1/\sqrt 2$, $f(0,-1/\sqrt 2)=3/2+1/\sqrt 2$.

Case B: $\lambda=1$. Then $6y-1=4y\Rightarrow 2y=1\Rightarrow y=1/2$. From constraint $x^2+2(1/4)=1\Rightarrow x^2=1/2\Rightarrow x=\pm 1/\sqrt 2$.

• $f(\pm 1/\sqrt 2,1/2)=1/2+3/4-1/2=3/4=0.75$.

Step 3 — Compare all candidate values

Absolute maximum: $f=3/2+1/\sqrt 2$ at $(0,-1/\sqrt 2)$.

Absolute minimum: $f=-1/12$ at $(0,1/6)$.

Answer