Science:Math Exam Resources/Courses/MATH105/April 2012/Question 07 (a)

MATH105 April 2012

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Question 07 (a)
Consider $\displaystyle f(x,y)=xye^{y}+{\frac {1}{2}}x^{2}-2.$ (a) Find all critical points of this function.

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
Both partial derivatives, ƒx and ƒy must vanish (equal 0) at a critical point.

Hint 2
Remember to use the product rule. Also, ey does not vanish for any value of y.

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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. To find the critical points of $f(x,y)=xye^{y}+{\frac {1}{2}}x^{2}-2$, we need to set both partial derivatives to 0: $\displaystyle f_{x}=ye^{y}+x=0$ and, using the product rule, $\displaystyle f_{y}=xe^{y}+xye^{y}=x(e^{y}+ye^{y})=0$ To solve the simultaneous system, we should first factor the equations as best we can (this makes it easier to find conditions that are necessary for the derivatives to vanish). Nothing can be done for the first equation so we leave it: $(1)\qquad f_{x}=ye^{y}+x=0$ For the second equation, we write: $(2)\qquad f_{y}=xe^{y}(1+y)=0$ It's easier to work with the equation above first. From above, we consider the possibilities that: x = 0, ey = 0 (this is impossible), or 1+y = 0, which happens when y = -1. Let's now use the possibilities above in the equation (1): ƒx = yey + x = 0. If x = 0, then yey + 0 = 0 implies yey = 0, so y = 0 (recall $e^{y}\neq 0$). Thus, $(0,0)$ is a critical point. If y = -1 then -e-1 + x = 0 implies x = e-1. Thus, $(e^{-1},-1)$ is a critical point.

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