You're looking for the extreme values of [tex]x^2+3y^2+13x[/tex] subject to the constraint [tex]x^2+y^2\le1[/tex].
The target function has partial derivatives (set equal to 0)
[tex]\dfrac{\partial(x^2+3y^2+13x)}{\partial x}=2x+13=0\implies x=-\dfrac{13}2[/tex]
[tex]\dfrac{\partial(x^2+3y^2+13x)}{\partial y}=6y=0\implies y=0[/tex]
so there is only one critical point at [tex]\left(-\dfrac{13}2,0\right)[/tex]. But this point does not fall in the region [tex]x^2+y^2\le1[/tex]. There are no extreme values in the region of interest, so we check the boundary.
Parameterize the boundary of [tex]x^2+y^2\le1[/tex] by
[tex]x=\cos u[/tex]
[tex]y=\sin u[/tex]
with [tex]0\le u<2\pi[/tex]. Then [tex]t(x,y)[/tex] can be considered a function of [tex]u[/tex] alone:
[tex]t(x,y)=t(\cos u,\sin u)=T(u)[/tex]
[tex]T(u)=\cos^2u+3\sin^2u+13\cos u[/tex]
[tex]T(u)=3+13\cos u-2\cos^2u[/tex]
[tex]T(u)[/tex] has critical points where [tex]T'(u)=0[/tex]:
[tex]T'(u)=-13\sin u+4\sin u\cos u=\sin u(4\cos u-13)=0[/tex]
[tex](1)\quad\sin u=0\implies u=0,u=\pi[/tex]
[tex](2)\quad4\cos u-13=0\implies\cos u=\dfrac{13}4[/tex]
but [tex]|\cos u|\le1[/tex] for all [tex]u[/tex], so this case yields nothing important.
At these critical points, we have temperatures of
[tex]T(0)=14[/tex]
[tex]T(\pi)=-12[/tex]
so the plate is hottest at (1, 0) with a temperature of 14 (degrees?) and coldest at (-1, 0) with a temp of -12.
