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## Homework Statement

I have a circular plate made of some metal and it is shaped like the equation [itex]x^2 + y^2 \leqslant 1 [/itex]

The temperature at any given point in this plate is given by the function [itex]T(x,y) = x^2 -x+2y^2[/itex]

I'm supposed to find the hottest and the coldest points.

## Homework Equations

If B^2 -AC < 0 and A < 0 ; C < 0 Then the point is a local maximum

If B^2 -AC < 0 and A > 0 ; C > 0 Then the point is a local minimum

## The Attempt at a Solution

first I try looking for critical points

[itex]\frac{\partial T}{\partial x} = 2x - 1

[/itex]

[itex]\frac{\partial T}{\partial y} = 4y [/itex]

the critical points are when the partial derivatives are equal to zero, so:

2x-1 = 0 → 2x = 1 → x = 1/2

4y = 0 → y = 0

so the point (1/2 , 0) is a critical point.

next is seeing if it's a minimum or maximum.

[itex]A = \frac{\partial^2 T}{\partial x^2} = 2

[/itex]

[itex]B = \frac{\partial^2 T}{\partial x \partial y } = 0[/itex]

[itex]C = \frac{\partial^2 T}{\partial y} = 2 [/itex]

since B^2 - AC < 0 and A > 0 ; C > 0

The point is local minimum.

So far ok

the answer in the book is

[itex](\tfrac{1}{2}, 0) [/itex]

is a minimum, which I did find

and

[itex](-\tfrac{1}{2},\tfrac{\sqrt{3}}{2}) , (-\tfrac{1}{2},-\tfrac{\sqrt{3}}{2})[/itex] are maxima

Now, my question is.

Where did the two maxima points come from?