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

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

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

The second partial derivatives are given by

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

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

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

The determinant of $H$ at the point $(1/2,1)$ is equal to $(-2e^{-2})\cdot \left(-{\frac {1}{2}}e^{-2}\right)-0\cdot 0=e^{-4}$ 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 ${\frac {\partial ^{2}f}{\partial x^{2}}}\left({\frac {1}{2}},1\right)=-2e^{-2}$ 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.