Global Max and Min Hi can someone please help me figure out how to start this question?
Temperature within a circular region $R = \{(x, y)| x^2 + y^2 \le 49\}$ is given by $T(x,y) = 4x^2 -4xy + y^2$
Find the global minimum and maximum temperatures in the region and the points where these extreme temperatures occur.
I'm supposed to use Lagrange multipliers and parametrization which I don't really understand. If someone can please help me get started I would really appreciate it.

Let $G(x,y) = x^2 + y^2 - 49$. At the interior points using critical points approach we have:
$T_x = 8x - 4y = 0$, and $T_y = 2y - 4x = 0$ both gives $y = 2x$. Thus. $T(x,2x) = 4x^2 - 4x(2x) + (2x)^2 = 0$. 
At the boundary, using Lagrange Multiplier we have:$T_x = \lambda G_x$, and $T_y = \lambda G_y$ gives:
$8x - 4y = \lambda 2x$, and 
$2y - 4x = \lambda 2y$.
Thus: $0 = 8x - 4y + 2(2y - 4x) = 2\lambda (x + 2y)$
So either $\lambda = 0$ or $x = -2y$.
Case 1: $\lambda = 0$,and $x \neq -2y$,  then $y = 2x$. Thus: $x^2 + (2x)^2 = 49 \to 5x^2 = 49 \to x = \pm \dfrac{7}{\sqrt{5}}$. So $y = \pm \dfrac{14}{\sqrt{5}}$. Thus: $T\left(\pm \dfrac{7}{\sqrt{5}},\pm \dfrac{14}{\sqrt{5}}\right) = 0$
Case 2: $x = -2y$, and $\lambda \neq 0$, then $(-2y)^2 +y^2 = 49 \to 5y^2 = 49 \to y = \pm \dfrac{7}{\sqrt{5}}$, and $x = \mp \dfrac{14}{\sqrt{5}}$. Thus: $T\left(\dfrac{7}{\sqrt{5}}, -\dfrac{14}{\sqrt{5}}\right) = \dfrac{784}{5} =  T\left(-\dfrac{7}{\sqrt{5}}, \dfrac{14}{\sqrt{5}}\right)$
Case 3: $\lambda = 0$, and $x = -2y$. So: $y = 2x = -4y$, and this gives: $y = 0 = x$. Thus: $T(0,0) = 0$.
In summary: $T_{min} = T(x,2x) = 0$, and $T_{max} = \dfrac{784}{5}$ achieved at $(x,y) = \left(\pm \dfrac{7}{\sqrt{5}}, \mp \dfrac{14}{\sqrt{5}}\right)$
Note: $T(x,y) = 4x^2 - 4xy + y^2 = (2x - y)^2 \geq 0$. This implies that $T_{min} = 0$ when $y = 2x$, and this is justified by the critical points method.
