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UPSC 2016 Maths Optional Paper 2 Q3b — Step-by-Step Solution

15 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 the relative maximum and minimum values of the function $f(x,y)=x^4+y^4-2x^2+4xy-2y^2$.

Technique

$\nabla f=0$ (adding the equations gives $y=-x$); Hessian test $D=f_{xx}f_{yy}-f_{xy}^2$; direct line-slices to resolve the degenerate $D=0$ case at the origin.

Solution

Find critical points ($\nabla f=0$), then classify with the second-derivative (Hessian) test using $D=f_{xx}f_{yy}-f_{xy}^2$. Where $D=0$ the test is inconclusive and we examine $f$ directly.

Step 1 — First-order conditions

Divide by $4$:

Add the two equations: $x^3+y^3=0\Rightarrow y=-x$ (real cube roots). Substitute $y=-x$ into the first: $x^3-x-x=0\Rightarrow x^3-2x=0\Rightarrow x(x^2-2)=0$, so $x=0,\ \pm\sqrt2$. With $y=-x$:

Step 2 — Second derivatives and the discriminant

At $(\sqrt2,-\sqrt2)$: $f_{xx}=12\cdot2-4=20$, $f_{yy}=20$, so $D=20\cdot20-16=384>0$ and $f_{xx}=20>0\Rightarrow$ relative minimum.

At $(-\sqrt2,\sqrt2)$: identical, $D=384>0$, $f_{xx}=20>0\Rightarrow$ relative minimum.

At $(0,0)$: $f_{xx}=-4$, $f_{yy}=-4$, $D=(-4)(-4)-16=16-16=0\Rightarrow$ test inconclusive.

Step 3 — Resolve $(0,0)$ directly

Examine $f$ along two lines through the origin:

• Along $y=x$: $f(t,t)=2t^4-2t^2+4t^2-2t^2=2t^4\ge0$, with $f(0,0)=0$ — $f$ increases away from $0$.
• Along $y=-x$: $f(t,-t)=2t^4-2t^2-4t^2-2t^2=2t^4-8t^2$, which is $<0$ for small $t\ne0$ ($\approx-8t^2$) — $f$ decreases away from $0$.

Since $f$ takes both signs in every neighborhood of $(0,0)$ (where $f=0$), the origin is a saddle point — neither a max nor a min.

Step 4 — Extreme values

At the two minima,

Answer