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I am asked to find the function that has the maximal area for a given length L when x runs from -a to a. I calculated the integral to be varied as follows:

$$ \int_{-a}^{a}\ y + \lambda \sqrt{1 + (\frac{dy}{dx})^2} dx $$

Where lambda is a Lagrange multiplier associated with the length constraint on the curve. After invoking Beltrami's identity I ended up with the following DE but could not figure out how to solve it:

$$ y + \lambda = C \sqrt{1+(\frac{dy}{dx})^2} $$

I would appreciate your help on this matter.

Edit: The answer given by Mathematica is as follows:

$$ \left\{\left\{y(x)\to \frac{1}{2} \left(C^2 e^{\frac{x}{C}-c_1}+e^{c_1-\frac{x}{C}}-2 p\right)\right\},\left\{y(x)\to \frac{1}{2} \left(C^2 e^{-c_1-\frac{x}{C}}+e^{c_1+\frac{x}{C}}-2 p\right)\right\}\right\} $$

Vesnog
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  • Hint: Do you have a suggestion, what the solution should look like? – gerw Jan 17 '14 at 19:03
  • Actually just heard from someone that it should be a circle but this was an exam question I was not able to answer. – Vesnog Jan 17 '14 at 19:54
  • So let $y$ be a function describing a circle and plug it into the DE. – gerw Jan 17 '14 at 20:24
  • @gerw Logical but I plugged this into Mathematica and it gave a different answer as far as I remember let me try again. – Vesnog Jan 17 '14 at 20:30

2 Answers2

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Actually after checking out "Mathematical Methods for Students of Physics and Related Fields" by Sadri Hassani and a website I concluded that the answer is a semicircle whose radius is given by $ \lambda^2 $

I am mirroring the treatment in the book by Hassani The treatment is as follows,

The function to be minimized is $$ \mathbf{L}[y] = \int_{-a}^{a}ydx $$ Under the following conditions and constraints $$ y(-a) = 0 = y(a) $$ $$ \mathbf{K}[y]=\int_{-a}^{a}\sqrt{1+y'^2}dx=L $$ Using Euler-Lagrange equation is more convenient in this case and it is given by: $$ \frac{\partial L}{\partial x}-\frac{d}{dt}\frac{\partial L}{\partial \dot{x}} + \lambda \left(\frac{\partial G}{\partial x} - \frac{d}{dt}\frac{\partial G}{\partial \dot{x}}\right)=0 $$ In our case $L=y$ and $G=\sqrt{1+y'^2}$, substituting thse into above equation gives $$ 1 + \lambda \frac{d}{dx} \frac{y'}{\sqrt{1+y'^2}}=0 $$ Integrating wrt x yields $$ x + \lambda \frac{y'}{\sqrt{1+y'^2}} = C_1 $$ Solve for $y'$ $$ y'=\pm \frac{C_1-x}{\sqrt{\lambda ^2-(C_1-x)^2}}=\frac{dy}{dx} $$

Multiply each side by $dx$ and integrate $$ y = \pm \sqrt{\lambda ^2-(C_1-x)^2}+C_2$$ Or $$ (x-C_1)^2+(y-C_2)^2=\lambda^2 $$ The unknowns are to be determined from the boundary conditions and the length constraint.

Vesnog
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  • OK, but the curve is not generally a semicircle. The curve is a circle, with the given line segment a chord, not necessarily a diameter. – Bill Kleinhans Jan 18 '14 at 00:06
  • Yeah $C_1$ and the other constants are to be determined but I could not manage to solve it through the use of Beltrami's identity and wonder the solution following from there. – Vesnog Jan 18 '14 at 09:43
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Approximate the curve by short line segments, and complete a polygon with the straight line segment from -a to a. Then use the fact that, given the length of the sides, the polygon with largest area can be inscribed in a circle. Not calculus of variations, but gets you quickly to the answer.

Bill Kleinhans
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  • Thanks for the answer. You are right about the circle after doing a bit of search I concluded that the largest area is indeed given by a circle but I need a calculus of variations proof to grasp the material. – Vesnog Jan 17 '14 at 20:47