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The same derivative here give different solutions if you solve it in different ways

  1. Jan 4, 2007 #1
    1) In the middle of a max min problem, i set the function that gave me the area that the problem requested. it's:

    [tex]A=\frac{\sqrt{2x-x^2}*(2x-1)}{2}[/tex] (but i can omit division by 2)

    [tex]A'=2\sqrt{2x-x^2}+\frac{(2-2x)(2x-1)}{2\sqrt{2x-x^2}}[/tex]

    [tex]A'=\frac{8x-4x^2-4x^2+2x+4x-2}{2\sqrt{2x-x^2}}[/tex]

    putting the derivative = 0,

    [tex]8x^2-14x+2=0[/tex], which gives the right result,

    [tex]x_\textrm{1,2}=\frac{7\pm \sqrt{33}}{8}[/tex].

    2) Now doing the square of the distances, i get:

    [tex]A=(2x-x^2)(2x-1)^2=8x^3+2x-8x^2-4x^4-x^2+4x^3[/tex]

    A'=0:

    [tex]8x^3-18x^2+9x-2=0[/tex]

    (not solvable with ruffini), should be equal to the result of the first derivative,

    [tex]x_\textrm{1,2}=\frac{7\pm \sqrt{33}}{8}[/tex].

    and in fact if i put the solutions of this first in the other, they work. but there is a third solution in the second, right? and however the equations are not the same, this worried me a lot on the problems i would get with square root of distances instead of pure distances.
    can anyone tell me why i don't get the same equation? thanks.
     
    Last edited: Jan 4, 2007
  2. jcsd
  3. Jan 4, 2007 #2

    Hootenanny

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    I believe you have neglected to differentiate your expression for A in the second part.
     
  4. Jan 4, 2007 #3
    sorry, what do you mean?
     
  5. Jan 4, 2007 #4

    Hootenanny

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    Sorry my mistake, I thought you had not differentiated. Indeed the additional solution to the expression is x = 1/2. You could probably discount this solution if you consider the physical situation, however, not having seen your question I cannot comment.
     
  6. Jan 4, 2007 #5

    HallsofIvy

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    I haven't checked but it is quite possible that the third solution to the cubic equation does not satisfy the original equation (with the square root) because it makes the root imaginary.
     
  7. Jan 4, 2007 #6

    D H

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    You have an error here. The correct solution is
    [tex]8x^3-18x^2+9x-1=(2x-1)(4x^2-7x+1)=0[/tex]

    The "area" is negative between 0 and 1/2, positive between 1/2 and 2. The square of the area has a zero at 1/2, and this is also a minimum in the square of the area.
     
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