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

MATH105 April 2013

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Question 04 (a)
Let $\displaystyle f(x,y)=xye^{-2x-y}$. Find all critical points of $f(x,y)$.

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Hint
What is true about the partial derivatives of a multi-variable function at critical points?

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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. The critical points of a multivariable function $f(x,y)$ are the values of $(x,y)$ where the following conditions are both simultaneously true: ${\begin{cases}{\frac {\partial f}{\partial x}}(x,y)=0\\{\frac {\partial f}{\partial y}}(x,y)=0\end{cases}}$ Since $\displaystyle f(x,y)=xye^{-2x-y}$, the above conditions become: ${\begin{cases}ye^{-2x-y}-2xye^{-2x-y}=0\\xe^{-2x-y}-xye^{-2x-y}=0\end{cases}}$ To solve these equations, we begin by factoring the left-hand sides, giving us ${\begin{cases}y(1-2x)e^{-2x-y}=0\\x(1-y)e^{-2x-y}=0\end{cases}}$ Since the exponential function is never zero, the equations above are equivalent to ${\begin{cases}y(1-2x)=0\quad &(1)\\x(1-y)=0.\quad &(2)\end{cases}}$ We now proceed to determine the values of $(x,y)$ such both Eq. (1) and (2) hold. The solutions to Eq. (1) are $y=0$ and $x=1/2$. We take each of these solutions and plug them into Eq. (2). If $y=0$, then Eq. (2) becomes: $x(1-0)=0\quad \rightarrow \quad x=0.$ Hence one of the critical points is $(x,y)=(0,0)$. If $x=1/2$, then Eq. (2) becomes: ${\frac {1}{2}}(1-y)=0\quad \rightarrow \quad y=1.$ So, another critical point is $(x,y)=(1/2,1)$. Therefore, the critical points of $f$ are ${\color {blue}\displaystyle (x,y)=(0,0),\left({\frac {1}{2}},1\right).}$

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