<s>[INST] Hottest and coldest points on a heated circular plate (use Lagrange multipliers) A circular plate given by the relationship $x^2 + y^2 \leq 1$ is heated according to the spatial temperature function $T(x,y) = 2x^2 + y^2-y$. Find the hottest and coldest point on the plate using Lagrange multipliers.

I tried to solve it but I got only one point (0,1) instead of 4 points .. I want to know how can I get 4 points ?? Is there any method to solve this problem?

 [/INST] I can provide a solution without using Lagrange multipliers.
$2x^2+y^2-y=2x^2+(y-\frac{1}{2})^2-\frac{1}{4}$
The coldest point is $\left(0,\frac{1}{2}\right)$, where the temperature is $-\frac{1}{4} $
The hottest point must be somewhere along the circle $x^2+y^2=1$, since the further away the a point is from $\left(0,\frac{1}{2}\right)$, the hotter it is. 
Let's use polar coordinates: $x=\cos\theta$, $y=\sin\theta$
We are then trying to maximize $2\cos^2\theta+\sin^2\theta-\sin\theta$
The derivative of this expression is $-\sin(2\theta)-\cos\theta$
The derivative equals zero at the following values of $\theta$: $\frac{\pi}{2}$, $\frac{3\pi}{2}$, $\frac{7\pi}{6}$, $\frac{11\pi}{6}$
Plugging these values of $\theta$ into the original equation, we can see that the hottest points are $\left(\pm\frac{\sqrt{3}}{2},-\frac{1}{2}\right)$ with temperature $\frac{9}{4}$. 
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