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Simple maximum/minimum application (temperature in a circular plate)

  1. Nov 14, 2011 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant 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

    3. 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?
     
  2. jcsd
  3. Nov 14, 2011 #2

    Dick

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    Homework Helper

    Those points are on the boundary of the set, x^2+y^2=1. They don't need to be critical points of T(x,y). You need to treat the boundary separately. Do you know how?
     
  4. Nov 14, 2011 #3

    probably not
    I guess the book hasn't mentioned this.
    or maybe it did by some other name...
     
  5. Nov 14, 2011 #4

    Ray Vickson

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    They are on the boundary x^2 + y^2 = 1, and at such points the derivatives of T do not need to vanish. In fact, you should look at the *constrained* problem max T(x,y), subject to x^2 + y^2 = 1. You can do this either by using a Lagrange multiplier method, or else by using the constraint to eliminate one of the two variables. For example, suppose we eliminate y by writing y^2 = 1-x^2. Plug that into T and get a simple one-dimensional problem, which you solve by setting the derivative to zero. (We can do that because we no longer have a constraint now.) That will give you x = -1/2; then y^2 = 3/4 will give you two y-values.

    Note: in general, be VERY careful about setting derivatives to zero when you have inequality constraints, because sometimes you fail to get the solution that way. The simplest example is max or min x, subject to 0 <= x <= 1. The min is at x = 0 and the max is at x = 1, but the derivative is nonzero at both those points!

    RGV
     
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