From part a), the critical points are
and the first partial derivatives are
![{\displaystyle {\frac {\partial f}{\partial x}}=y(1-2x)e^{-2x-y}=0\quad {\frac {\partial f}{\partial y}}=x(1-y)e^{-2x-y}=0}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/d85ab1129e21cbe1d91037f997039ff22ee0f6cf)
The second partial derivatives are given by
![{\displaystyle {\begin{aligned}{\frac {\partial ^{2}f}{\partial x^{2}}}&=y(-2)e^{-2x-y}+y(1-2x)(-2)e^{-2x-y}\\&=-2ye^{-2x-y}(1+1-2x)\\&=-4ye^{-2x-y}(1-x)\\&=4ye^{-2x-y}(x-1)\\\\{\frac {\partial f}{\partial x\partial y}}&=y(1-2x)e^{-2x-y}\\&=(1-2x)e^{-2x-y}+y(1-2x)(-1)e^{-2x-y}\\&=(1-2x)e^{-2x-y}(1+y)\\&=(1-y-2x+2xy)e^{-2x-y}\\\\{\frac {\partial ^{2}f}{\partial y^{2}}}&=x(-1)e^{-2x-y}+x(1-y)(-1)e^{-2x-y}\\&=xe^{-2x-y}(-1-1+y)\\&=xe^{-2x-y}(y-2)\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/c02a7520396f24b3044990271fbbde7e80a5709a)
To classify the critical points, we need to compute the Hessian matrix,
, of the function
:
![{\displaystyle H(x,y)=\left[{\begin{array}{cc}{\frac {\partial ^{2}f}{\partial x^{2}}}&{\frac {\partial ^{2}f}{\partial x\partial y}}\\{\frac {\partial ^{2}f}{\partial x\partial y}}&{\frac {\partial ^{2}f}{\partial y^{2}}}\end{array}}\right]=\left[{\begin{array}{cc}4y(x-1)e^{-2x-y}&(1-y-2x+2xy)e^{-2x-y}\\(1-y-2x+2xy)e^{-2x-y}&x(y-2)e^{-2x-y}\end{array}}\right].}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/6164d9375a6db845f27b1bbf4f3add3ab02e133b)
Evaluating
at the critical point
gives
![{\displaystyle H(0,0)=\left[{\begin{array}{cc}0&1\\1&0\end{array}}\right].}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/cb3ef3a60f62fd364cd2b0a57afe5849c3637b23)
The determinant of
at the point
is equal to
, which is less than zero. Hence the point
is a saddle point.
Evaluating
at the critical point
gives
![{\displaystyle H\left({\frac {1}{2}},1\right)=\left[{\begin{array}{cc}-2e^{-2}&0\\0&-{\frac {1}{2}}e^{-2}\end{array}}\right].}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/b2f1d2f5dc30df50fdca33ac42d40248ba487f5a)
The determinant of
at the point
is equal to
which is greater than zero and so
is not a saddle point of
and must be either a local max or min. Since
is less than zero, the point
is a local maximum of
.
Therefore, (x,y) = (0,0) is a saddle point and (x,y) = (1/2,1) is a local maximum.