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

MATH105 April 2014

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Question 04 (a)
Let $T(x,y)={\frac {1}{9}}y^{3}+x^{2}-2xy+6x-6y$. Find all critical points of $T(x,y)$ and classify each as a local maximum, local minimum or saddle point.

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Hint
Calculate the critical points. At a critical point $T_{x}$ and $T_{y}$ are equal to zero. Then use the second derivative test to check if the critical point is a maximum, minimum or saddle point.

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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. 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.

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