Question

A flat circular plate has the shape of the region x squared plus y squared less than or equals 1x2+y2≤1. the​ plate, including the boundary where x squared plus y squared equals 1x2+y2=1​, is heated so that the temperature at the point left parenthesis x comma y right parenthesis(x,y) is upper t left parenthesis x comma y right parenthesist(x,y)equals=x squared plus 3 y squared plus one third xx2+3y2+13x. find the temperatures at the hottest and coldest points on the plate.

Answer 1

You're looking for the extreme values of $x^2+3y^2+13x$ subject to the constraint $x^2+y^2\le1$.

The target function has partial derivatives (set equal to 0)

$\dfrac{\partial(x^2+3y^2+13x)}{\partial x}=2x+13=0\implies x=-\dfrac{13}2$

$\dfrac{\partial(x^2+3y^2+13x)}{\partial y}=6y=0\implies y=0$

so there is only one critical point at $\left(-\dfrac{13}2,0\right)$. But this point does not fall in the region $x^2+y^2\le1$. There are no extreme values in the region of interest, so we check the boundary.

Parameterize the boundary of $x^2+y^2\le1$ by

$x=\cos u$

$y=\sin u$

with $0\le u$. Then $t(x,y)$ can be considered a function of $u$ alone:

$t(x,y)=t(\cos u,\sin u)=T(u)$

$T(u)=\cos^2u+3\sin^2u+13\cos u$

$T(u)=3+13\cos u-2\cos^2u$

$T(u)$ has critical points where $T'(u)=0$:

$T'(u)=-13\sin u+4\sin u\cos u=\sin u(4\cos u-13)=0$

$(1)\quad\sin u=0\implies u=0,u=\pi$

$(2)\quad4\cos u-13=0\implies\cos u=\dfrac{13}4$

but $|\cos u|\le1$ for all $u$, so this case yields nothing important.

At these critical points, we have temperatures of

$T(0)=14$

$T(\pi)=-12$

so the plate is hottest at (1, 0) with a temperature of 14 (degrees?) and coldest at (-1, 0) with a temp of -12.