Question

The function $f(x,y)=x^{2}y-3xy+2y+x$ has

• A. No local extremum
• B. One local maximum but no local minimum
• C. One local minimum but no local maximum
• D. One local minimum and one local maximum

Answer 1

The correct option is A No local extremum

Using necessary conditions for maxima-minima
$\frac{\partial f}{\partial x}=2xy-3y+1=0$ ...(1)
$\frac{\partial f}{\partial y}=x^2-3x+2=0$ .... (2)
On solving (1) & (2) we get $x=1,y=1$ & $x=2,y=-1$
So $P_1(1,1)and{P}_2(2,-1)$ are the critical points
Now, At $(1,1)$ and $(2,-1)$; value of $rt-s^2<0$
i.e., $[\frac{{\partial }^{2}f}{\partial x^2}\times \frac{{\partial }^{2}f}{\partial y^2}]-{\left[\frac{{\partial }^{2}f}{\partial x\partial y}\right]}^2<0$
$\Rightarrow$ Both$P_1$ & $P_2$ are saddle points
So no local extremum