Why F is the Way it is for Classical Mechanics Problem

[stunner5000pt]

of the classical kind

All i need here is an explanation as why F is the way it is
A person wished to fence off a maximum area for hi8s dog by attachin a flexible fence of length L to the side of his house whose width is 2a. What should the slope of the fence be?
Assume L > 2a
well then the action
A[y] = \int_{-a}^{a} y(x) dx is a maximum
subject to constant L where L = \int_{-a}^{a} \sqrt{1+y'^2} dx
If L[y,y'} = constant then \delta L = 0
and A[y,y'] = constant at its maximum
then
\delta A + \lambda \delta L = 0 where
\lambda = \frac{[A]}{[L]}
we also want that \delta \int_{-a}^{a} F(y,y',x) dx = 0
where F = y + \lambda \sqrt{1+y'^2}
thats the problem, Why is F the way it is ? Maybe the attached diagram helps...

 

[stunner5000pt]

can anyone help out?

is it because f takes on the shape of half a unit circle?

Or will any function do?

 

[lightgrav]

maximizing the Area provided the "y" term,
using all the fence material provided the sqrt term.
Whether you put the lambda multplier with the sqrt term
or with the y term was your choice, back in step 4 ...