Curve enclosing the maximum area

Solution 1:

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 $\lambda$

Solution 2:

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.

Solution 3:

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.