Solving path optimisation problem without euler Lagrange equation

Question

If a curve given by y=f(x) starts at (0,0) and ends at (x1,y1), then find the function f(x) such that the area under the curve from x=0 to x=x1 is maximum under the constraint that the total length of curve between these is fixed and is equal to l. I know this question can be solved using euler Lagrange equation but I was wondering if there is some other simple and intuitive method for solving this question.

Answer 1

I will present a quite intuitive and simple method which just uses high school calculus to solve this question.

Let us suppose that the required curve is given by $y=f(x)$. Now consider a particular point $x,f(x)$ and also its immediate neighbours $$x-dx,f(x-dx)$$ and $$x+dx,f(x+dx)$$ Since we are considering a small element, therefore we can assume that it is nearly straight. So we assume that the line joining the middle point to its left neighborhood point has inclination $A$ and that of the line joining the middle point to its right neighborhood is $A+dA$. Further we see that $$tan(A)= f'(x)=$$ {$f(x)-f(x-dx)$}$/dx $

and $$tan(A+dA)= f'(x+dx)= $${$f(x+dx)-f(x)$}$/dx$

as well as

$f''(x)=$

{$f'(x+dx)-f'(x)$}$/dx$

$ = d(tanA)/dx = sec²A(dA/dx)$

This is a bit weird way of representation but does not matter much. Maybe someone else can write it in a better way. It is just my style.

Now I will do a mischievous trick that I will stretch the middle point by a very small amount $dh$. I will assume that $dh <<<< dx,(dx)²$. So the middle point now becomes $x,f(x)+dh$. This slight upliftment will cause an infinitesimal change in arc length. By using correct approximations we can see that the small change in arc length is

$dl = sinA dh$(due to left) $- sin(A+dA) dh$(due to right)

$=${$sin(A)-sin(A+dA)$}$dh$ $$= -d(sinA)dh$$

Also, there will be infinitesimal change in area under curve which can be given by $da = dx dh $ ( by adding the area of 2 new triangles formed both having base $dh$ and height $dx$). We can clearly see that though both da and dl are infinitesimal but da/dl is non zero finite number which is a function of x only as dh gets cancelled.

Now here is my argument:-

"The ratio da/dl must have same value at all values of x".

This is because if we suppose this ratio has different values, say 1 and 2 at 2 different points then it would be possible to slightly increase the height at one point and decrease at the other point in such a way that there is net 0 change in total arc length, yet there will be non zero change in area under curve, that is it can be made to increase and decrease. This contradicts with our initial condition that $y=f(x)$ is the optimum path for maximum area under curve. Therefore it is necessary that $dh/dl = C$ where $C$ is a constant and not a function of x.

Therefore we get $$d(sinA)/dx = C$$ So, $$ sinA= Cx+ D$$ where $D$ is another constant. Now

$ sinA = tanA/ (1+tan²A)$^$(1/2) $ and $ tan A = dy/dx$. Now when we will solve this differential equation ( I leave it to the OP), we will get an equation of circle with 3 constants. 2 of these can be calculated by the boundary conditions, that is where the curve starts and ends, and one will be found by the total arc length of curve.