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**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.

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