Lagrange multiplier Calculus of variations

Question

considering the problem
$$\text{max} \int_0^1 v\ dx$$
$$\int_0^1 \sqrt{1+v'^2}\ dx=1$$
I would like to solve it with the Lagrange multipliers.
the Lagrangian is
$$\mathcal L(v,v')=\int_0^1v-\lambda \sqrt{1+v'^2}dx=\int_0^1 l(v,v')dx$$
and now I would try to use the Eulero-Lagrange equation
$$\frac{d}{dt}\frac{\partial l}{\partial v'}=\frac{\partial l}{\partial v}$$
but after this passage I obtain a very complicated differential equation
$$\left( \frac{v'}{\sqrt{1 + (v')^2}}\right)'=\lambda$$
while the correct solution should be
$$\frac{v''}{\sqrt{1+(v')^2}^3}=-\lambda ^{-1}$$
Have I made a mistake in the procedure? Or the answer is wrong?

Answer 1

First, I see two errors: according to your Lagrangian, you should have a RHS of 1 and a coefficient of $-\lambda$ on your LHS. After that, actually take your derivative and simplify and you'll see cancellations leaving the solution. Note that following the product rule, you have $$\left( \frac{v'}{\sqrt{1+v'^2}} \right)' = v''\frac{1}{\sqrt{1+v'^2}} - v'\frac{v'v''}{(1+v'^2)^{3/2}}.$$