Science:Math Exam Resources/Courses/MATH105/April 2011/Question 07/Solution 1

The derivative of $f$ with respect to $x$ is

$f_x(x,y)=(1/x)+2bx+3ay$.

Also, the derivative of $f$ with respect to $y$ is

$f_y(x,y)=(2/y)+3ax$.

For a critical point at $(1,3)$, we require that $f_x(1,3)=f_y(1,3)=0$. Using the above two equations, this gives us the following system to solve for $a$ and $b$:

1. $f_x(1,3)=(1/1)+2b(1)+3a(3)=1+2b+9a=0$
2. $f_y(1,3)=(2/3)+3a(1)=2/3+3a=0$.

Equation 2 tells us that $a=-2/9$.

Substituting this value for $a$ into equation 1 tells us that $1+2b+9(-2/9)=1+2b-2=2b-1=0$. Therefore, we have that $b=1/2$.

This critical point is a saddle point. To show this, we solve for the second derivatives $f_{xx}$, $f_{xy}$ and $f_{yy}$.

First,

$f_{xx}(x,y)=-(1/x{)}^2+2b$.

Therefore,

$f_{xx}(1,3)=-(1)+2(1/2)=0$.

Second,

$f_{yy}(x,y)=-2(1/y{)}^2$.

Therefore,

$f_{yy}(1,3)=-2/9$

Third,

$f_{xy}(x,y)=3a$.

Therefore,

$f_{xy}(1,3)=-2/3$.

We can classify the critical points by looking at the formula,

$D=f_{xx}{f}_{yy}-(f_{xy}{)}^2.$

If $D<0$ then the critical point is a saddle point. If $D>0$ then we look to the sign of $f_{xx}$. If $f_{xx}>0$ then we have a local minimum and if $f_{xx}<0$ then we have a local maximum. For our problem,

$D(1,3)=0(-\frac{2}{9})-{(-\frac{2}{3})}^2=-\frac{4}{9}<0.$

Therefore since $D<0$ the critical point is a saddle point.

Advanced:

Using the second partial derivatives of a function $f(x)$ we can define a general Hessian matrix as

$\left[\begin{array}{cc}{f}_{xx}& f_{xy}\\ f_{yx}& f_{yy}\end{array}\right]$

which for our specific example is

$\left[\begin{array}{cc}0& -2/3\\ -2/3& -2/9\end{array}\right]$.

The eigenvalues of this matrix are given by the solving the formula $(-\lambda )(-2/9-\lambda )-4/9=0$ for $\lambda$.

That is,

${\lambda }^2+(2/9)\lambda -4/9=0$.

Solving gives us the answer

$\lambda =(1/9)(\pm \sqrt{37}-1)$.

Because one eigenvalue is positive and one eigenvalue is negative, we conclude that the Hessian matrix is indefinite, and thus the point $(1,3)$ is a saddle point.