The derivative of with respect to is
- .
Also, the derivative of with respect to is
- .
For a critical point at , we require that . Using the above two equations, this gives us the following system to solve for and :
- .
Equation 2 tells us that .
Substituting this value for into equation 1 tells us that . Therefore, we have that .
This critical point is a saddle point. To show this, we solve for the second derivatives , and .
First,
- .
Therefore,
- .
Second,
- .
Therefore,
Third,
- .
Therefore,
- .
We can classify the critical points by looking at the formula,
If then the critical point is a saddle point. If then we look to the sign of . If then we have a local minimum and if then we have a local maximum. For our problem,
Therefore since the critical point is a saddle point.
Advanced:
Using the second partial derivatives of a function we can define a general Hessian matrix as
which for our specific example is
- .
The eigenvalues of this matrix are given by the solving the formula for .
That is,
- .
Solving gives us the answer
- .
Because one eigenvalue is positive and one eigenvalue is negative, we conclude that the Hessian matrix is indefinite, and thus the point is a saddle point.