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
:
![{\displaystyle f_{x}(1,3)=(1/1)+2b(1)+3a(3)=1+2b+9a=0}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/d72cb701ab5f2505ebc471502e412f4c843dcf69)
.
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,
![{\displaystyle f_{yy}(1,3)=-2/9}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/f73463c987041b7319e48436044d62d52f6e26f6)
Third,
.
Therefore,
.
We can classify the critical points by looking at the formula,
![{\displaystyle D=f_{xx}f_{yy}-(f_{xy})^{2}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/b778afa9d9545b37e61bd0fae66d286dbcbddba7)
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,
![{\displaystyle D(1,3)=0\left(-{\frac {2}{9}}\right)-\left(-{\frac {2}{3}}\right)^{2}=-{\frac {4}{9}}<0.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/b9c904a76088f3afce99b856335d628c6eaa5472)
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
![{\displaystyle {\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/9aa0cf6b3cb3ae5907d36746a4e1b6ddea57cece)
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.