(adsbygoogle = window.adsbygoogle || []).push({}); 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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

**Physics Forums | Science Articles, Homework Help, Discussion**