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Question

# Let T = g(x, y) be the temperature at the point (x, y) on the ellipse $x = 2 \sqrt { 2 } \cos t , \quad y = \sqrt { 2 } \sin t , \quad 0 \leq t \leq 2 \pi,$ and suppose that$\frac { \partial T } { \partial x } = y , \quad \frac { \partial T } { \partial y } = x.$Suppose that T = xy - 2. Find the maximum and minimum values of T on the ellipse.

Solution

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We have:

\begin{aligned} T&=g(x,y)=xy-2 \\ g\left(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2} \right)&=\dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{2}}{2}-2=\dfrac{1}{2}-2=-\dfrac{3}{2} \\ g\left(-\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2} \right)&=\left(-\dfrac{\sqrt{2}}{2} \right)\cdot\dfrac{\sqrt{2}}{2}-2=-\dfrac{1}{2}-2=-\dfrac{5}{2} \end{aligned}

We get that:

• Maximum: $T=-\frac{3}{2}$
• Minimum: $T=-\frac{5}{2}$

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