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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.

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up vote 1 down vote accepted

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.

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Think about how we approach such problems in single variable calculus. There our boundaries were only endpoints of our interval but the same kind of idea applies. We had to check the extrema inside the interval and then check the function values on the boundary (i.e. the endpoints of our interval). The same kind of idea applies here. We check inside via the standard techniques; however in several variables, things are a little bit more delicate since boundaries can be a bit weird. There you implement Lagrange multipliers. With that said, I'll set up the problem but I'll leave you to solve it.

First, let's find the extrema inside our region. The way we do this is, as per usual, to set $T_x = 0$ and $T_y = 0$ simultaneously to find which $(x,y)$ pairs solve this. This is nothing more than the first derivative test. I trust you can take it from here.

As for the boundary, let's employ Lagrange multipliers. If $g$ represents our level curve for our boundary, then $\nabla T = \lambda \nabla g$. ($\nabla T$ is nothing more than $(T_x,T_y)$ and similarly for $g$). Our boundary is given by nothing more than $g(x,y) = x^2+y^2 = 49$ so $g_x = 2x$ and $g_y = 2y$. We also have that $T_x = 8x-4y$ and $T_y = -4x+2y$.

Setting $T_x = \lambda g_x$ and $T_y = \lambda g_y$, we get the following equations: $8x-4y = 2\lambda x$ and $-4x+2y = 2\lambda y$. Multiplying the second equation by $2$ gives $-8x+4y = 4\lambda y$. Adding this to our first equation gives $(8x-4y)+(-8x+4y) = 2\lambda x+4\lambda y$. The left hand side is zero so $0 = 2\lambda(x+2y)$.

This says either that $\lambda = 0$ or $x=-2y$. If $\lambda = 0$, then $T_x = 0$ and $T_y = 0$ (Is this possible?). Can you take it from here?

For some worked examples, see Paul's Online Notes.

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We use Lagrange multipliers. Let, $F(x,y) = 4x^2 - 4xy + y^2 + \lambda(x^2 + y^2 - 49)$. Thus, $$ \begin{cases} \dfrac{\partial F}{\partial x} = 8x - 4y + 2\lambda x = 0 \qquad (1)\\ \dfrac{\partial F}{\partial y} = 8y - 4x + 2\lambda y = 0 \qquad(2)\\ \dfrac{\partial F}{\partial \lambda} = x^2 + y^2 -49= 0 \qquad (3) \end{cases} $$ From equations (1) and (2), we have, $$ \dfrac{2x - 4y}{y} = \lambda = \dfrac{2y -4x}{x} \quad \Rightarrow \quad x^2 = y^2 $$ Thus, from equation (3), $2x^2 = 49 \ \Rightarrow \ x = \pm \frac{7}{\sqrt{2}}$ and $y = \pm \frac{7}{\sqrt{2}}$ The points of maximum and minimum are $(-\frac{7}{\sqrt{2}},-\frac{7}{\sqrt{2}})$ and $(\frac{7}{\sqrt{2}},\frac{7}{\sqrt{2}})$.

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Bear in mind that $\sin^2 + \cos^2 = 1$ always, so the frontier of $R$ can be parametrized by $\gamma: [0, 2\pi[ \rightarrow R$, $\gamma(t) = (7 \cos t, 7 \sin t)$. So, you can search for critical points of $$T(\gamma(t)) = 4(7\cos t)^2 - 4(7\cos t)(7\sin t) + (7\sin t)^2$$ just like in single variable calculus.

And to look for critical points in the interior of $R$, just find the points that satisfy $$\frac{\partial T}{\partial x} = \frac{\partial T}{\partial y} = 0$$ Do each part separately, and gather all the information at the end. Ok?

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$\partial T(x,y)/\partial x=8x-4y=0.$

$\partial T(x,y)/\partial y=2y-4x=0.$

Solving this: $y=2x$

Hence: $x^2+4x^2\le 49$ or $x^2\le49/5$

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