UPSC 2013 Maths Optional Paper 2 Q3c — Step-by-Step Solution

10 marks · Section A

Maxima and minima of single-variable functions · Real Analysis · asked 3× in 13 yrs · Read the full method →

Question

Let $f(x,y)=y^{2}+4xy+3x^{2}+x^{3}+1$ . At what points will $f(x,y)$ have a maximum or minimum?

Standard multivariable Hessian test.

Strategy. Find critical points by setting $\nabla f=0$ ; classify using the second-derivative (Hessian) test.

f_x=6x+4y+3x^{2}=0,\qquad f_y=2y+4x=0.

From $f_y=0$ : $y=-2x$ . Substitute into $f_x=0$ :

6x+4(-2x)+3x^{2}=0\;\Longrightarrow\;-2x+3x^{2}=0\;\Longrightarrow\;x(3x-2)=0.

So $x=0$ (giving $y=0$ ) or $x=\tfrac{2}{3}$ (giving $y=-\tfrac{4}{3}$ ).

Two critical points: $P_1=(0,0)$ and $P_2=\bigl(\tfrac{2}{3},-\tfrac{4}{3}\bigr)$ .

$f_{xx}=6+6x,\quad f_{yy}=2,\quad f_{xy}=4$ .

Hessian discriminant:

H=f_{xx}f_{yy}-(f_{xy})^{2}=2(6+6x)-16=12x-4.

At $P_1=(0,0)$ : $H=-4<0$ . Saddle point. Neither max nor min.

At $P_2=(\tfrac{2}{3},-\tfrac{4}{3})$ : $H=12\cdot\tfrac{2}{3}-4=8-4=4>0$ , and $f_{xx}=6+6\cdot\tfrac{2}{3}=10>0$ . Local minimum.

f(\tfrac{2}{3},-\tfrac{4}{3})=\tfrac{16}{9}-\tfrac{32}{9}+\tfrac{12}{9}+\tfrac{8}{27}+1.

First three: $\tfrac{16-32+12}{9}=-\tfrac{4}{9}=-\tfrac{12}{27}$ .

$-\tfrac{12}{27}+\tfrac{8}{27}=-\tfrac{4}{27}$ . Plus 1: $\tfrac{23}{27}$ .

f_{\min}^{\text{local}}=\tfrac{23}{27}.

As $x\to\infty$ (with any $y$ , e.g. $y=0$ ), $f(x,0)=3x^{2}+x^{3}+1\to\infty$ . So $f$ is unbounded above, and there is no maximum.

\boxed{\;\text{Local minimum at }(2/3,-4/3)\text{ with value }23/27;\text{ saddle point at }(0,0);\text{ no maximum.}\;}