Curve enclosing the maximum area the curve of fixed length $l$ that joins the points $(0,0)$ and $(1,0)$ lies above the $x-axis$ and encloses the maximum area between itself and the $x-axis$, is a segment of 


*

*A straight line

*A parabola

*An ellipse

*A circle


I don't know exactly how to solve it but it seems Circle is the right Answer
 A: The above problem once again:
A curve of fixed length l lies above the x-axis, joins the points (0,0) and (1,0), and encloses the maximum area between itself and the x-axis is ---  a circle.
The constraints on l:
1) l>1 (obvious)
2) l=< π/2 
Let's look at constraint 2:
2) : As l approaches π/2 the centre of the circle moves up, from below the x-axis( negative y-values) along  a line parallel to.the x-axis through (1/2,0).
For l= π/2 the centre of the circle is at (1/2,0), maximal area is a semi circle.
Assume l>π/2 .
Problem: As before, this time with l>π/2.
Find the curve that encloses the maximal area between itself and the x-axis.
Comments welcome.
A: Here l is the fixed perimeter of plane curve passing through two given points (0,0) and (1,0). Let S be the area enclosed by the plane curve and x-axis 
Then, we are maximize $S=\int _0^1 y\;dx$ with boundary conditions y(0)=y(1)=0 subject to constraint $\int_0^1(1+y'^2)^\frac{1}{2}\;dx=l$
Let $F(x,y,y')=y+ \lambda(1+y'^2)^\frac{1}{2} $
Where $\lambda$ is the lagrange multiplier . Then the required extremal satiesfied the Euler's equation 
$\frac{\partial F}{\partial y}-\frac{d}{dx}(\frac{\partial F}{\partial y'})=0\\
\Rightarrow 1-\frac{d}{dx}\{\frac{\lambda y'}{(1+y'^2)^\frac{1}{2}}\}=0\\
Integrating \;w\;r\;to\;x, x-\{\frac{\lambda y'}{(1+y'^2)^\frac{1}{2}}\}=a\\
\Rightarrow (x-a)^2=\{\frac{\lambda^2 y'^2}{(1+y'^2)}\}\\
\Rightarrow \frac{1+y'^2}{y'^2}=\frac{\lambda^2}{(x-a)^2}\\
\Rightarrow y'^2=\frac{(x-a)^2}{\lambda^2-(x-a)^2}\\
\frac{dy}{dx}=\pm \frac{(x-a)}{(\lambda^2-(x-a)^2)^\frac{1}{2}}\\
Integrating, \; y=b\pm\{\lambda^2-(x-a)^2\}^\frac{1}{2}\\
\Rightarrow (y-b)^2=\lambda^2-(x-a)^2\\
\Rightarrow (x-a)^2+(y-b)^2=\lambda^2$
this is equation of a circle with radius $\lambda$ 
A: A. little more to the above problem, maybe of interest.
(x-a)^2 + (y-b)^2 = (Lambda)^2.
Centre of the circle at (a,b) , radius Lambda.
1)Curve passes through (0,0) ,
2) Curve passes through (1,0),
3)Length of curve is l.
3 equations for 3 unknowns, OK.
The centre of any circle passing through 2 points lies on the perpendicular bisector of the segment joining the 2 points, in this case (0,0) and (1,0), I.e. a line parallel to the y-axis passing through (1/2,0).
So the x coordinate of the centre of the circle : a = 1/2.
Check:
Setting x=0,y=0 and x=1,y=0 in the equation of the above circle gives 
1) a^2 +b^2 = (Lambda)^2 and 
2) (1-a)^2 + b^2 = (Lambda)^2
which gives a = 1/2.
Now to b and Lambda ( radius ).
Consider the line parallel to the y-axis through (1/2,0). The centres of all possible circles passing through (0,0) and (1,0) are on this line.
Pick a point on this line below the x -axis, call it M, and join it to (0,0) and (1,0): a segment of a circle.
Call the angle subtended  by straight line joining M and (0,0) and M and (1/2,0) Phi.
1) Lambda × sin (Phi) = 1/2
2) 2× Lambda × (Phi) = l ( length)
2 equations for two unknowns,  Lambda (radius) and Phi, angle subtended. 
( b = - Lambda × cos(Phi) )
Dividing eq. 1) by eq.2) gives:
3) sin (Phi) = (1/l) × (Phi)
Left hand side is a sin fct., right hand side a linear fct. of (Phi). 
There is, with the constraints on l mentioned below, exactly one solution 0 < (Phi) <= π/2, I.e. there is exactly one point of intersection of the sin curve with the straight line
(See reference below).
Constraints for l:
1) l >= 1 , the length of the curve must be greater than the length of the straight line joining (0,0) and (1,0) (shortest distance) 
2) l =< π/2 , largest l for centre of circle located at (1/2,0).
Comments are welcome.
Ref.
Thomas Mueller, Mathe-Seiten, Isoperimetrisches Problem.
A: By reflecting the curve across the $x$-axis we can make it in to a closed loop, which doubles both the length and the area. Thus this is equivalent to the usual isoperimetric problem (maximize area inside a loop of fixed length) with the constraint of reflection symmetry. Since the circle is the unconstrained optimum and also has the reflection symmetry, it provides the solution to your problem: the curve is a semicircle.
