# Let g(x,y) = xye^{ x^2 y^2} . 1. Find all critical points. 2. Classify the critical points...

## Question:

Let {eq}g(x,y) = xye^{-x^2-y^2} {/eq}.

1. Find all critical points.

2. Classify the critical points found above.

## Critical Points and their Classification:

Given a function {eq}f(x,y) {/eq} that is differentiable with respect to {eq}x {/eq} and {eq}y {/eq}, let {eq}(a,b) {/eq} be critical points of {eq}f(x,y) {/eq} such that {eq}f_x(a,b) = 0 {/eq} and {eq}f_y(a,b) = 0 {/eq}. Let {eq}D {/eq} be equal to {eq}f_{xx}(a,b)f_{yy}(a,b) - [f_{xy}(a,b)]^2 {/eq}. If {eq}D < 0 {/eq}, then there is a saddle point. If {eq}D > 0 {/eq}, then the point is a local minimum if {eq}f_{xx}(a,b) > 0 {/eq} or a local maximum if {eq}f_{xx}(a,b) < 0 {/eq}.

Solve for the system of equations:

{eq}\displaystyle g_x = ye^{x^2y^2}(1+2x^3y^2) = 0 \\ \displaystyle g_y = xe^{x^2y^2}(1+2x^2y^3) = 0 \\ {/eq}

If x = 0:

{eq}\displaystyle y(1) = 0 \\ \displaystyle x(1) = 0 \\ \displaystyle x = y = 0 \\ {/eq}

If {eq}\displaystyle 2x^3y^2 = -1 {/eq}:

{eq}\displaystyle xe^{x^2y^2}(-2x^3y^2+2x^2y^3) = 0 \\ \displaystyle x(x^2y^2)(y - x) = 0 \\ \displaystyle y = x \\ \displaystyle 2x^5 = -1 \\ \displaystyle x = y = \left(\frac{-1}{2}\right)^{\frac{1}{5}} \\ {/eq}

Get the second partial derivatives:

{eq}\displaystyle g_{xx} = y(e^{x^2y^2})[(6x^2y^2) + (1+2x^3y^2)(2xy^2)] \\ \displaystyle g_{yy} = x(e^{x^2y^2})[(6x^2y^2) + (1+2x^2y^3)(2x^2y)] \\ \displaystyle g_{xy} = e^{x^2y^2}(1+2x^3y^2)+ y(1+2x^3y^2)(2x^2y)e^{x^2y^2} + ye^{x^2y^2}(4x^3y) \\ {/eq}

At (0,0):

{eq}\displaystyle g_xx = 0 \\ \displaystyle g_yy = 0 \\ \displaystyle g_xy = 1 \\ \displaystyle D = -1 \\ {/eq}

At {eq}\displaystyle \left(\left(\frac{-1}{2}\right)^{\frac{1}{5}}, \left(\frac{-1}{2}\right)^{\frac{1}{5}} \right) {/eq}:

{eq}\displaystyle g_xx = -5.327 \\ \displaystyle g_yy = -5.327 \\ \displaystyle g_xy = -3.551 \\ \displaystyle D = 15.76 \\ {/eq}

The first point is a saddle point, while the second point is a maximum.