Answer:
We have the coldest value of temperature
. and the hottest value is
.
Step-by-step explanation:
We need to take the derivative with respect of x and y, and equal to zero to find the local minimums.
The temperature equation is:

Let's take the partials derivatives.
So, we can find the critical point (x,y) of T(x,y).




The critical point is (3/4,0) so the temperature at this point is:
Now, we need to evaluate the boundary condition.

We can solve this equation for y and evaluate this value in the temperature.

Now, let's find the critical point again, as we did above.
Evaluating T(x,y) at this point, we have:
Now, we can see that at point (3/4,0) we have the coldest value of temperature
. On the other hand, at the point
we have the hottest value of temperature, it is
.
I hope it helps you!