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,

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

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

${\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}}$

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.