← 2023 Paper 1

UPSC 2023 Maths Optional Paper 1 Q3b — Step-by-Step Solution

15 marks · Section A

Maxima and Minima of Multi-Variable Functions (Unconstrained) · Calculus · Read the full method →

Question

Justify whether $(0,0)$ is an extreme point for the function $f(x,y)=2x^4-3x^2y+y^2$.

Technique

Hessian test fails (zero determinant). Probe along curves $y=\alpha x^2$ — the natural family because the cubic-in-$y$ structure of $f$ (treating $x^2$ as a parameter) reveals the sign behaviour.

Solution

Step 1 — Find critical point and Hessian at $(0,0)$.

$f_x=8x^3-6xy.$ $f_y=-3x^2+2y.$ At $(0,0)$: $f_x=f_y=0$ — critical point.

$f_{xx}=24x^2-6y,\,f_{yy}=2,\,f_{xy}=-6x.$ At $(0,0)$: $f_{xx}=0,\,f_{yy}=2,\,f_{xy}=0$. Hessian determinant: $0\cdot 2-0=0$. Hessian test inconclusive.

Step 2 — Approach $(0,0)$ along curves $y=\alpha x^2$.

Let $y=\alpha x^2$ for various $\alpha\in\mathbb R$. Substitute:

The sign of $f$ near $(0,0)$ along this approach:

• For $\alpha<1$: $(\alpha-1)<0$ and $(\alpha-2)<0$, so product $>0$. $f>0$.
• For $\alpha=1$: $f=0$ (along this parabola, $f$ vanishes identically).
• For $1<\alpha<2$: $(\alpha-1)>0,\,(\alpha-2)<0$, so product $<0$. $f<0$.
• For $\alpha>2$: both positive, product $>0$. $f>0$.

Step 3 — Conclusion.

In any neighbourhood of $(0,0)$, $f$ takes both positive and negative values:

• $f>0$ along $y=0$ (where $f=2x^4\ge 0$, $>0$ for $x\ne 0$).
• $f<0$ along $y=1.5\,x^2$ (where $f=x^4(0.5)(-0.5)=-x^4/4<0$ for $x\ne 0$).

Therefore $(0,0)$ is not a local extremum (neither minimum nor maximum). It is a saddle-like critical point.

Answer