Setting both partial derivatives to 0 gives a single critical point at
, which does fall inside the unit disk.
At this point, the value of the derivative of the Hessian matrix is
while the value of the second-order partial derivative with respect to
is
This means the critical point is the site of a local minimum, so this is the coldest point on the plate with a temperature of
.
The hottest point on the plate must then be found on the boundary. Let
and
, so that
Then the boundary of the plate (the circle
) is a function of a single variable
considered over
. Differentiating once gives
You'll find that
attains three extrema on the interval
, with relative maxima at
and
and a relative minimum at
(and
, if you want to include that).
We already found our minimum on the inside of our plate - which you can verify to have a lower temperature than at the points given by
- and we find two maxima at
and
, each giving a maximum temperature of
.
Converting back to Cartesian coordinates, these points correspond to the points
.