# Thread: Help with min/max when D = 0

1. ## Help with min/max when D = 0

f(x,y) = e ^ (-x^2-y^2)

at point (0,0), D = 0. Therefore the normal min/max test fails.

The problem asks to classify the critical point at the origin by visualizing the surface z = f (x,y)

when:

f(x,0) = e^(-x^2) which is 1 / e(x^2) which would be POSITIVE
f(-x,o) = POSITIVE
f(0,y) = POSITIVE
f(o,-y) = POSITIVE

Shouldn't this mean this is a local/global minimum since as we move away from the point the function increases?

However, the book says (0,0) would be a global maximum. Can someone explain?

2. Originally Posted by messianic
f(x,y) = e ^ (-x^2-y^2)

at point (0,0), D = 0. Therefore the normal min/max test fails.

The problem asks to classify the critical point at the origin by visualizing the surface z = f (x,y)

when:

f(x,0) = e^(-x^2) which is 1 / e(x^2) which would be POSITIVE
f(-x,o) = POSITIVE
f(0,y) = POSITIVE
f(o,-y) = POSITIVE

Shouldn't this mean this is a local/global minimum since as we move away from the point the function increases?

However, the book says (0,0) would be a global maximum. Can someone explain?
$\displaystyle f_x=-2xe^{-(x^2+y^2)}$
$\displaystyle f_y=-2ye^{-(x^2+y^2)}$
$\displaystyle f_{xx}=2e^{-(x^2+y^2)}(2x^2-1)$
$\displaystyle f_{yy}=2e^{-(x^2+y^2)}(2y^2-1)$
$\displaystyle f_{xy}=4xye^{-(x^2+y^2)}$

At $\displaystyle (0,0)$, $\displaystyle f_{xx}=f_{yy}=-1$ and $\displaystyle f_{xy}=0$ so $\displaystyle f_{xx}f_{yy}-f_{xy}^2=1$. This means we have either a local min or a local max (not a saddle point). But since $\displaystyle f_{xx}(0,0)$ is negative, we can conclude that it is a local max.