The problem I'm having trouble with is this;
The temperature at all points in the disk $x^2 + y^2 {\leq} 1$ is given by $T = (x+y)e^{-x^2-y^2}$. Fine the minimum and maximum temperatures at the points of the disk.
So I know we can start at $x^2 + y^2 = 1$, which means $y$ or $x = 0$
I found the gradient to be ${\nabla}f(x,y) = ((x+y)(-2xe^{-x^2-y^2}) + (1+y)e^{-x^2-y^2}, (x+y)(-2ye^{-x^2-y^2}) + (1+x)(e^{-x^2-y^2}))$
Though I don't see how this could be much help.
Can someone guide me through this problem?
Thanks very much in advance.
One approach to solve this is to find stationary points (or critical points) in the interior of the disk first, and afterwards look at the boundary of the disk by itself.
To find stationary points in the interior, solve the gradient equal to zero for $(x,y)$, and check if they're within the disk. That is, solve $$\nabla T(x,y) = \begin{bmatrix} \frac{dT}{dx}\\\frac{dT}{dy}\end{bmatrix}= \begin{bmatrix} 0\\0\end{bmatrix}. $$ The system is
$$ \nabla T(x,y) = \begin{bmatrix} -e^{-x^2-y^2}(2x^2+2xy-1)\\-e^{-x^2-y^2}(2y^2+2xy-1)\end{bmatrix} = \begin{bmatrix} 0\\0\end{bmatrix}$$ which implies $$\begin{bmatrix} 2x^2+2xy-1\\2y^2+2xy-1\end{bmatrix}=\begin{bmatrix} 0\\0\end{bmatrix} $$ and solving yields $(x,y) = \left(\frac12,\frac12\right)$ or $(x,y)=\left(-\frac12,-\frac12\right)$.
To check the boundary, you can parametrize it by
$$(x(t),y(t)) = (\cos t,\sin t) $$ where $t\in[0,2\pi)$. Again, find stationary points (with respect to $t$), by solving $$\frac{d}{dt}T(x(t),y(t))=0.$$
We have that $$T(x(t),y(t)) = (\cos t +\sin t)e^{-(\cos^2 t + \sin^2 t)} = (\cos t +\sin t)e^{-1} $$ and hence its derivative $$\frac{d}{dt}T(x(t),y(t)) = (-\sin t + \cos t)e^{-1} = 0$$ which implies $$-\sin t + \cos t = 0 $$ which occurs when $t=\frac\pi 4+p\pi$ and $p\in\mathbb{Z}$. Since we're restricted to $t\in[0,2\pi)$, we only have the options $p=0$ or $p=1$, that is $t=\frac14\pi$ or $t =\frac54\pi$. We now have four candidates for maxima and minima namely \begin{align} (x,y)&=\left(\frac12,\frac12\right)\\ (x,y)&=\left(-\frac12,-\frac12\right)\\ (x,y)&=\left(x\left(\frac14\pi\right),y\left(\frac14\pi\right)\right)=\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\\ (x,y)&=\left(x\left(\frac54\pi\right),y\left(\frac54\pi\right)\right)=\left(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right). \end{align} Check $T$ in all these points and you know what minimum and maximum is.