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 $$\left( \frac{v'}{\sqrt{1+v'^2}} \right)' = v''\frac{1}{\sqrt{1+v'^2}} - v'\frac{v'v''}{(1+v'^2)^{3/2}}.$$