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

MATH105 April 2012

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

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[show] Hint 1
Both partial derivatives, ƒ_x and ƒ_y must vanish (equal 0) at a critical point.

[show] Hint 2
Remember to use the product rule. Also, e^y does not vanish for any value of y.

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[show] Solution
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, e^y = 0 (this is impossible), or 1+y = 0, which happens when y = -1. Let's now use the possibilities above in the equation (1): ƒ_{x = ye^y + x = 0.} If x = 0, then ye^y + 0 = 0 implies ye^y = 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.