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The critical points of a multivariable function are the values of where the following conditions are both simultaneously true:
![{\displaystyle {\begin{cases}{\frac {\partial f}{\partial x}}(x,y)=0\\{\frac {\partial f}{\partial y}}(x,y)=0\end{cases}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/3bccfdb571805eafb018ea8511da0c59b761bba6)
Since , the above conditions become:
![{\displaystyle {\begin{cases}ye^{-2x-y}-2xye^{-2x-y}=0\\xe^{-2x-y}-xye^{-2x-y}=0\end{cases}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/b941061f6e347142b0f484924520b1a8bac56abc)
To solve these equations, we begin by factoring the left-hand sides, giving us
![{\displaystyle {\begin{cases}y(1-2x)e^{-2x-y}=0\\x(1-y)e^{-2x-y}=0\end{cases}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/1e73b29b7283e90ebb4fbb5d4c6886ca965124d9)
Since the exponential function is never zero, the equations above are equivalent to
![{\displaystyle {\begin{cases}y(1-2x)=0\quad &(1)\\x(1-y)=0.\quad &(2)\end{cases}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/ac47cdac3f0ee62e8a8e08808c6f78a60e826f06)
We now proceed to determine the values of such both Eq. (1) and (2) hold. The solutions to Eq. (1) are and . We take each of these solutions and plug them into Eq. (2).
If , then Eq. (2) becomes:
![{\displaystyle x(1-0)=0\quad \rightarrow \quad x=0.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/3ea7d97b9e121eb662e51038fd8c4389339674d2)
Hence one of the critical points is .
If , then Eq. (2) becomes:
![{\displaystyle {\frac {1}{2}}(1-y)=0\quad \rightarrow \quad y=1.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/2b36df6d2142e859a5b1233d9077a48d1089a34e)
So, another critical point is .
Therefore, the critical points of are
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