Find the local max and min values and saddle points. (Partial derivatives)

Question

f(x,y) = xye^{-(x^2 + y^2)/2}

Answer 1

f(x,y) = xy/exp[(x²+y²)/2]

 

Find the gradient:

 

∂f/∂x = {y exp[(x²+y²)/2] - x²y exp[(x²+y²)/2]}/exp(x²+y²)

         = y(1-x²)/exp[(x²+y²)/2]

 

∂f/∂y = x(1-y²)/exp[(x²+y²)/2]

 

So the gradient is 0 if x(1+y)(1-y)=0 and y(1+x)(1-x)=0

 

We get (0,0), (1,1), (1,-1), (-1,1), and (-1,-1).

 

We test these with a Hessian.

 

∂²f/∂x² = {-2xy exp[(x²+y²)/2] - xy(1-x²)exp[(x²+y²)/2]}/exp(x²+y²)

           = xy(x²-3)/exp[(x²+y²)/2]

 

∂²f/∂y² = 2xy(y²-3)/exp[(x²+y²)/2]

 

∂²f/∂x∂y = {(1-y²)exp[(x²+y²)/2] - x²(1-y²)exp[(x²+y²)/2]}/exp(x²+y²)

             = (1-x²)(1-y²)/exp[(x²+y²)/2]

 

det[Hf(x,y)] = [4x²y²(x²-3)(y²-3) - (1-x²)(1-y²)]/exp(x²+y²)

                  =

 

Plug the points in:

At (0,0), ∂²f/∂x² = 0.

At (1,1) and (-1,-1), ∂²f/∂x² = -2/e < 0.

At (-1,1) and (1,-1), ∂²f/∂x² = 2/e > 0.

 

At (0,0), det[Hf(x,y)] = -1/e² < 0

 

At (1,1), (-1,-1), (-1,1), and (1,-1), det[Hf(x,y)] = 36/e² > 0

 

Saddle points: f = 0 at (0,0).

 

Maxima: f = 1/e at (1,1) and (-1,-1)

 

Minima: f = -1/e at (-1,1) and (1,-1)