Hottest and coldest points on a heated circular plate (use Lagrange multipliers)

Question

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?

Answer 1

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 $\left(\pm\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$.

Answer 2

The extremal temperatures are taken either in the interior $D$ or on the boundary $\partial D$ of the unit disk. Writing $$T(x,y)=2x^2+\left(y-{1\over2}\right)^2-{1\over4}$$ we see that the graph of $T$ is a paraboloid with its apex at $P_1=\bigl(0,{1\over2}\bigr)\in D$, and there are no other stationary points of $T$ in $D$. Take note of $T(P_1)=-{1\over4}$.

For $\partial D$ we use the parametrization $\phi\mapsto(\cos\phi,\sin\phi)$ and then have to consider the pullback $$\hat T(\phi)=2\cos^2\phi+\sin^2\phi-\sin\phi={9\over4}-\left(\sin\phi+{1\over2}\right)^2\ .$$ The function $\hat T$ is maximal (namely $={9\over4}$) at the points where $\sin\phi=-{1\over2}$.

This leads to the candidates $P_2=\bigl(\cos{7\pi\over6},\sin{7\pi\over6}\bigr)$ and $P_3=\bigl(\cos{11\pi\over6},\sin{11\pi\over6}\bigr)$. Take note of $T(P_2)=T(P_3)={9\over4}$.

The function $\hat T$ is minimal at the points where $\sin\phi=1$. This leads to the candidate $P_4=(0,1)$ with $T(P_4)=\hat T\bigl({\pi\over2}\bigr)=0$.

Summing it up we can say that it is coldest (namely $-{1\over4}$) at $P_1\in D$ and hottest (namely ${9\over4}$) at the points $P_2$, $P_3\in\partial D$.

Answer 3

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 $\frac{9}{4}$.