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?


Write
$$T (x,y) = 2 x^2 + y^2 - y = 2 x^2 + \left(y - \frac{1}{2}\right)^2 - \frac{1}{4} \geq -\frac{1}{4}$$
Thus, the coldest point is $(0,\frac{1}{2})$, where the temperature is $-\frac{1}{4}$.
The hottest point must be on the circle $x^2 + y^2 = 1$, which is the domain of interest's boundary.
Introduce the Lagrangian
$$\mathcal{L} (x,y,\lambda) = T (x,y) + \lambda \left(x^2 + y^2 - 1\right)$$
Taking the partial derivatives with respect to $x,y,\lambda$ and finding where they vanish, we get
$$(2+\lambda) \, x = 0, \qquad (1+\lambda) \, y = \frac{1}{2}, \qquad x^2+y^2=1$$
If $\lambda = -2$, then we get $y = -\frac{1}{2}$. Since $x^2 + y^2 = 1$, we have the two hot points
$$\left(\pm\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$$
where the temperature is $\dfrac{9}{4}$. If $\lambda \neq -2$, then
$$x = 0, \qquad y = \frac{1}{2(1+\lambda)}, \qquad x^2+y^2=1$$
Hence, $(1+\lambda)^2 = \frac{1}{4}$, which gives us $\lambda \in \{-\frac{3}{2},-\frac{1}{2}\}$. If


*

*$\lambda = -\frac{3}{2}$, we get the hot point $(0,-1)$, where the temperature is $2$. 

*$\lambda = -\frac{1}{2}$, we get the point $(0,1)$, where the temperature is $0$. 


Since $2 < \frac{9}{4}$, the hottest points are $\left(\pm\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$.
