# Finding function that maximizes ratio of area to length

I'm new to variational analysis, so I need someone to check, if I'm going in the right direction.

Let's say I need to find a curve $y(x)$ with $y(0) = 1$ and $y(1) = 0$ that maximizes ratio of area (enclosed by a curve and $x$-axis) to length of this curve. In other words, I need to maximize $$\int_0^1F(y,y')dx,\text{ where } F(y,y') = \frac{y}{\sqrt{1+y'^2}}.$$ Thus Euler-Lagrange equation simplifies to this: $$F_y - \frac{d}{dx}F_{y'} = 0,$$ $$F_y - F_{yy'}y' - F_{y'y'}y'' = 0,\text{ since }F_{xy'} = 0.$$ Now multiplying it by $y'$ simplifies to $$\frac{d}{dx}(F-y'F_{y'}) = 0,$$ $$F-y'F_{y'} = C,C\text{ is constant. }$$ Plugging our $F(y, y')$ into the last equation gives $$\frac{y}{\sqrt{1+y'^2}} - y'\left(-\frac{1}{2}\frac{2yy'}{\sqrt{(1+y'^2)^3}}\right) = C,$$ $$\frac{y(1+y'^2)+yy'^2}{\sqrt{(1+y'^2)^3}} = C,$$ $$\frac{2yy'^2 + y}{\sqrt{(1+y'^2)^3}} = C.$$ Now how to solve this would be my next question, but first I need to know, if variations-wise I'm correct. Thanks.

To find maximum area for a given length their ratio is taken as the Lagrange multiplier. (primes are understood as differentiation with respect to x.)

$$\dfrac{\int y dx}{\int \sqrt{(1+y^{'2} }dx}= \lambda$$

where we take the constant as ratio of unevaluated indefinite integrals.

We cannot divide the quantities right under the integral sign!

Considering object and constraint together we find constrained maximum of

$$\int( y- \lambda \sqrt{(1+y^{'2} }) \, dx$$ where the Lagrangian is:

$$F(x,y,y^{'}) = (y- \lambda \sqrt{(1+y^{'2} })$$

This when applied to Euler-Lagrange Equation results in constant curvature circles

$$y y''/(1+y^{'2})^{3/2} = \lambda$$

The integrand should pass through (0,1) and (1,0), allowing determination of integration constants.

EDIT1:

If you had been asked hypothetically to find in an unconstrained problem to maximize

$$\int y(x) \cos\phi\; dx$$

where $\phi$ is tangent slope then for that problem the ODE you found out is correct.