Science:Math Exam Resources/Courses/MATH105/April 2013/Question 04 (b)/Solution 1

From part a), the critical points are $(x,y)=(0,0),(\frac{1}{2},1)$ and the first partial derivatives are

$\frac{\partial f}{\partial x}=y(1-2x)e^{-2x-y}=0\frac{\partial f}{\partial y}=x(1-y)e^{-2x-y}=0$

The second partial derivatives are given by

$\begin{array}{rl}\frac{{\partial }^{2}f}{\partial x^2}& =y(-2)e^{-2x-y}+y(1-2x)(-2)e^{-2x-y}\\ & =-2y{e}^{-2x-y}(1+1-2x)\\ & =-4y{e}^{-2x-y}(1-x)\\ & =4y{e}^{-2x-y}(x-1)\\ \\ \frac{\partial f}{\partial x\partial y}& =y(1-2x)e^{-2x-y}\\ & =(1-2x)e^{-2x-y}+y(1-2x)(-1)e^{-2x-y}\\ & =(1-2x)e^{-2x-y}(1+y)\\ & =(1-y-2x+2xy)e^{-2x-y}\\ \\ \frac{{\partial }^{2}f}{\partial y^2}& =x(-1)e^{-2x-y}+x(1-y)(-1)e^{-2x-y}\\ & =x{e}^{-2x-y}(-1-1+y)\\ & =x{e}^{-2x-y}(y-2)\end{array}$

To classify the critical points, we need to compute the Hessian matrix, $H$, of the function $f(x,y)$:

$H(x,y)=\left[\begin{array}{cc}\frac{{\partial }^{2}f}{\partial x^2}& \frac{{\partial }^{2}f}{\partial x\partial y}\\ \frac{{\partial }^{2}f}{\partial x\partial y}& \frac{{\partial }^{2}f}{\partial y^2}\end{array}\right]=\left[\begin{array}{cc}4y(x-1)e^{-2x-y}& (1-y-2x+2xy)e^{-2x-y}\\ (1-y-2x+2xy)e^{-2x-y}& x(y-2)e^{-2x-y}\end{array}\right].$

Evaluating $H$ at the critical point $(x,y)=(0,0)$ gives

$H(0,0)=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right].$

The determinant of $H$ at the point $(0,0)$ is equal to $(0\cdot 0)-(1\cdot 1)=-1$, which is less than zero. Hence the point $(0,0)$ is a saddle point.

Evaluating $H$ at the critical point $(x,y)=(1/2,1)$ gives

$H(\frac{1}{2},1)=\left[\begin{array}{cc}-2e^{-2}& 0\\ 0& -\frac{1}{2}{e}^{-2}\end{array}\right].$

The determinant of $H$ at the point $(1/2,1)$ is equal to $(-2e^{-2})\cdot (-\frac{1}{2}{e}^{-2})-0\cdot 0=e^{-4}$ which is greater than zero and so $(1/2,1)$ is not a saddle point of $f$ and must be either a local max or min. Since $\frac{{\partial }^{2}f}{\partial x^2}(\frac{1}{2},1)=-2e^{-2}$ is less than zero, the point $(1/2,1)$ is a local maximum of $f$.

Therefore, (x,y) = (0,0) is a saddle point and (x,y) = (1/2,1) is a local maximum.