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

Compute the derivatives:

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

$T_{x x} (x, y) = 2, T_{x y} (x, y) = - 2, T_{y y} (x, y) = \frac{2}{3} y .$

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

$2 x - 2 y + 6 = 0$

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

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

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

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