Find minimizer of the functional $ l(u)= \int \limits _{-1} ^1 u(t) \mathbb d t $ with $u(-1)=u(1)=0 $ subject to $g(u)=\int \limits _{-1} ^1 \sqrt{1+u'(t)} \mathbb d t=π $.
I solved it using Lagrange's equations and I found $u(t)=\sqrt{\lambda ^2 -(t+c)^2 }+c$.
First I started by $l^*=l- \lambda g$ then I used the Euler-Lagrange equation ($l_u -\frac{d}{dt}l_u'=0$) or first integral ($l-u'l_u'=c$).
My problem is how to find value of $c$.
Using the initial conditions $u(-1)=u(1)=0$, just impose $\sqrt {\lambda ^2 - (-1 + c)^2} + c = \sqrt {\lambda ^2 - (1 + c)^2} + c = 0$. The first equality gives $c=0$. Next, using $c=0$, the second equality will give $\lambda = \pm 1$.