Question

Determine all the relative minimum and maximum values, and saddle points of the function g defined by g(x,y) = x³ - 3x +3xy²

Answer 1

$f_x=3x^2-3+3y^2$

$f_{xx}=6x$

$f_y=6xy$

$f_{yy}=6x$

$f_{xy}=6y$

Critical points

$f_x=3x^2-3+3y^2=0$

$f_y=6xy=0$

We can set them equal to each other

$3x^2-3+3y^2=6xy$

$x^2-2xy+y^2=1$

$(x-y)^2=1$

x-y=1

y=x-1....[i]

Solving fy=0 gives

Either x=0 , or y=0

When x=0,y=−1⟹(x,y)=(0,−1)

When y=0,x=1⟹(x,y)=(1,0)

We know

D=f_xxf_yy−[fxy]²

We require D<0 for the point (x,y) to be a saddle point

When (x,y)=(0,−1)

$D=6(0)6(0)-(6(-1))^2=-36$

straightforward result for a confirmed saddle point.

When (x,y)=(1,0)

$D=6(1)6(1)-(6(0))^2=36>0$

This is not saddle point. As D>0 and f_xx>0, so it is relative minimum.

Conclusion: The saddle point is (x,y)=(0,−1), relative minimum is (1,0)