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

MATH105 April 2013

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Question 04 (b)
Let $\displaystyle f(x,y)=xye^{-2x-y}$. Classify each critical point you found as a local maximum, a local minimum, or a saddle point of $f(x,y)$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
To classify critical points of a multi-variable function, we need to use the Hessian, the matrix of second partial derivatives.

Hint 2
Evaluate the Hessian at the critical points, take the determinant and see what the resulting value tells you about the type of critical point.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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.

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