Science:Math Exam Resources/Courses/MATH105/April 2014/Question 04 (a)/Solution 1

Compute the derivatives:

$T_{x}(x,y)=2x-2y+6\quad \quad T_{y}(x,y)={\frac {1}{3}}y^{2}-2x-6;$

$T_{xx}(x,y)=2,\;\quad T_{xy}(x,y)=-2,\;\quad T_{yy}(x,y)={\frac {2}{3}}y.$

Set $T_{x}(x,y)=0$ and $T_{y}(x,y)=0$ to find critical points:

$2x-2y+6=0\;$

${\frac {1}{3}}y^{2}-2x-6=0.$

From the first equality we get $x=y-3$. Plugging this into the second equality, we get ${\frac {1}{3}}y^{2}-2y=0$. Solving it gives $y=0$ and $y=6$. This yields the critical points $(x,y)=(-3,0)$, and $(x,y)=(3,6)$.

For $(-3,0)$, $T_{xx}=2$, $T_{yy}={\frac {2}{3}}y=0$ and $T_{xy}=-2$, so we have a saddle point because $T_{xx}T_{yy}-T_{xy}^{2}=-4<0$.

For $(3,6)$, $T_{xx}=2$, $T_{yy}={\frac {2}{3}}y=4$, and $T_{xx}\cdot T_{yy}-T_{xy}^{2}=2\cdot {\frac {2}{3}}y-(-2)^{2}=2\cdot 4-(-2)^{2}=4>0$ (and thus it could be a local max or local min). Then, because $T_{xx}=2>0$, we conclude it is a local minimum.