Optimal String Shape Problem So here is the problem I am working on,
Given a curve of length L connecting the points (0,1) and (1,0) find an expression for the equation of the curve that minimizes the area underneath it.
In other words, given that:
$$ y(0) = 1, y(1) = 0$$
minimize
$$ \int_0^1{y(x)} dx $$
subject to:
$$ \int_0^1{\sqrt{1 + (y')^2} } dx = l $$
So my initial strategy was to do this problem using functional lagrange multipliers (I don't know if thats a real thing but my intuition is that it should work) whereas I make an optimization function
$$ \omega = \int_0^1{y(x)} dx + \lambda(\int_0^1{\sqrt{1 + (y')^2} } dx - l) $$
Which can be rewritten as:
$$ \omega = \int_0^1{[y(x) + \lambda \sqrt{1 + (y')^2} - \lambda l]} dx  $$
We now compute the functional gradient of it as
$$ \nabla \omega  = 0 $$
--->
$$ \frac{\delta \omega}{\delta y} = 0 $$
$$  \frac{\delta \omega}{\delta \lambda} = 0 $$
Which results in 
$$ 1  - \lambda \frac{y''}{(1 +  (y')^2)^{\frac{3}{2}}} = 0 $$
$$ \int_0^1{\sqrt{1 + (y')^2} } dx = l $$
Now i'm not sure what to do at this stage... Help would be appreciated!! (My problem is I was trying to eliminate $\lambda$ and solve for y as a single differential equation which didn't end up happening

As per Alex's Answer
The first equation yields that
$$ 1  - \lambda \frac{y''}{(1 +  (y')^2)^{\frac{3}{2}}} = 0 $$
We can set $y' = u$
$$ 1  - \lambda \frac{u'}{(1 +  (u)^2)^{\frac{3}{2}}} = 0 $$
Which yields 
$$ \frac{1}{\lambda} = \frac{1}{(1 +  (u)^2)^{\frac{3}{2}}} u'$$
We can hit this with the Leibniz Chain Rule. To find
$$ \int \frac{1}{\lambda} dx = \frac{u}{(1+u^2)^{\frac{1}{2}}} $$
Furthermore that yields
$$ \left(\int \frac{1}{\lambda} dx \right)^2 (1 + u^2) = u^2 $$ 
Which allows us to solve for $u$ as 
$$ \frac{\int \frac{1}{\lambda} dx}{\left( 1 - \left(\int \frac{1}{\lambda} dx \right)^2 \right)^{\frac{1}{2}} }  = u$$
Then it follows that 
$$ y = \int  \frac{\int \frac{1}{\lambda} dx}{\left( 1 - \left(\int \frac{1}{\lambda} dx \right)^2 \right)^{\frac{1}{2}} } dx $$ 
If we add in constants
$$ y = \int  \frac{\int \frac{1}{\lambda} dx+C_1}{\left( 1 - \left(\int \frac{1}{\lambda} dx + C_1 \right)^2 \right)^{\frac{1}{2}} } dx + C_2 $$ 
Now we have the constraints that 
$$ y(0) = 1, y(1) = 0, \int_0^1{\sqrt{1 + (y')^2} } dx = l $$
Given that $\lambda$ is a function i'm not sure how to proceed.
 A: If you remove the triangle with corners (0,1),(1,0),(0,0),and rotate by $\pi/4$  the problem is identical to Dido's,in variational calculus in standard text books.
A: How about asking wolfram alpha for the solution for the first equation?
As you can see, there is a complex factor, however, if we make the term under the square root negative, we obtain two different real solutions. If we plug in the boundary conditions, we get two equations coupling the constants with $\lambda$. Now, using the constraint could make the solution unique?
But maybe - I did not do the calculations - you get stuck in the computations above...
EDIT: I get - if I did not make any mistakes - using only the first possible solution of the ode:
$$c_{2}=i\sqrt{(c_{1}\lambda+1)^2-\lambda^2}$$ and
$$c_{1}=\tfrac{\lambda^{3}-3\lambda\pm\sqrt{-\lambda^{6}+6\lambda^{4}-\lambda^{2}}}{4\lambda^{2}}.$$ You see, both constants can be eliminated and in the solution we finally have only the dependence on $\lambda$. This can now be inserted in the constraint (i.e. the length constraint), and you can determine $\lambda$. However, it is a very ugly computation :-).
