(3) (math 105 final, 2013) find the critical points of $f(x,y)=xye^{-2x - y}$

Explanation:

Step1: Find first - order partial derivatives

First, find $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$. Using the product rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = xy$ and $v = e^{-2x - y}$. $\frac{\partial f}{\partial x}=y\cdot e^{-2x - y}+xy\cdot(- 2)e^{-2x - y}=ye^{-2x - y}(1 - 2x)$ $\frac{\partial f}{\partial y}=x\cdot e^{-2x - y}+xy\cdot(-1)e^{-2x - y}=xe^{-2x - y}(1 - y)$

Step2: Set partial derivatives equal to zero

Set $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$. From $\frac{\partial f}{\partial x}=ye^{-2x - y}(1 - 2x)=0$, we have $y = 0$ or $x=\frac{1}{2}$. From $\frac{\partial f}{\partial y}=xe^{-2x - y}(1 - y)=0$, we have $x = 0$ or $y = 1$.

Case 1: If $y = 0$, substituting into $\frac{\partial f}{\partial y}=0$, when $y = 0$, $x$ can be any real - number. But substituting $y = 0$ into $\frac{\partial f}{\partial x}=0$, we get critical points when $x$ is arbitrary. Let's consider the intersection of the two zero - set conditions. If $y = 0$, from $\frac{\partial f}{\partial y}=xe^{-2x - y}(1 - y)=0$, when $y = 0$, $x = 0$ gives a critical point $(0,0)$. If $x=\frac{1}{2}$, substituting into $\frac{\partial f}{\partial y}=xe^{-2x - y}(1 - y)=0$, we get $1 - y = 0$, so $y = 1$. The critical point is $(\frac{1}{2},1)$. If $x = 0$, substituting into $\frac{\partial f}{\partial x}=ye^{-2x - y}(1 - 2x)=0$, when $x = 0$, $y$ can be any real - number. But considering both conditions, $(0,0)$ is a valid critical point. If $y = 1$, substituting into $\frac{\partial f}{\partial x}=ye^{-2x - y}(1 - 2x)=0$, we get $1 - 2x = 0$, so $x=\frac{1}{2}$.

Answer:

The critical points of $f(x,y)$ are $(0,0)$ and $(\frac{1}{2},1)$