minimize $$\int_0^1 y(t)\sqrt{1+(y'(t))^2} \,dt$$
such that y(0) = y(1) = 0 and $$\int_0^1 \sqrt{1+(y'(t))^2} \,dt = 2$$
EDIT: I tried using the Euler Lagrange relations and came up with this equation:
$\frac{d}{dt} \frac{yy'}{\sqrt{1+(y')^2}}$ = $\sqrt{1+(y')^2}$ which leads to
$yy'= (1+(y')^2)t$
I don't know where to go from here.
EDIT 2: I followed up with notes on Beltrami identity and here is what I got and where I am currently stuck at:
$y\sqrt{1+y'^2} - yy'^2 = a\sqrt{1+y'^2}$
$y = a\sqrt{1+y'^2}$
$\frac{y}{a} = \sqrt{1+y'^2}$
$$ \int_0^1 \frac{y}{a} \,dt = \int_0^1 \sqrt{1+(y'(t))^2} \,dt = 2$$
I am not sure where to go from $$ \int_0^1 \frac{y}{a} \,dt = 2$$