I am trying to find the extremal (if it exists) of the isoperimetric problem with the functional: $I[x] = \int_0^1(\dot{x}^2+2x)dt$ subject to the conditions:
$\int_0^1 2x dt = 3$, $x(0) = -1$, $x(1) = 4$
I use the Mayer equations to derive a value of an extremal in terms of Lagrange multipliers. When I solve for these multipliers, I get two possible values and therefore two extremums. Is this possible? Is it possible to have two extremums for this problem? I've only ever been used to solving for one.