Solving Max-Min Problem with Square of Distances

[Born2Perform]

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

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

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

A'=\frac{8x-4x^2-4x^2+2x+4x-2}{2\sqrt{2x-x^2}}

putting the derivative = 0,

8x^2-14x+2=0, which gives the right result,

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

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

A=(2x-x^2)(2x-1)^2=8x^3+2x-8x^2-4x^4-x^2+4x^3

A'=0:

8x^3-18x^2+9x-2=0

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

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

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.

 

[Hootenanny]

I believe you have neglected to differentiate your expression for A in the second part.

 

[Born2Perform]

I believe you have neglected to differentiate your expression for A in the second part.

sorry, what do you mean?

 

[Hootenanny]

sorry, what do you mean?

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.

 

[HallsofIvy]

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.

 

[D H]

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

A=(2x-x^2)(2x-1)^2=8x^3+2x-8x^2-4x^4-x^2+4x^3

A'=0:

8x^3-18x^2+9x-2=0

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

can anyone tell me why i don't get the same equation? thanks.

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.