1. ## Maximization problem

Find the functions x(t) and y(t) that maximize $\int_0^1(4x-y^2)dt$ subject to $\dot x = y, x(0)=1, x(1)=2$

So I formed the Hamiltonian:

$H=4x-y^2 + \lambda y$

And took the partials:

$\frac{\partial H}{\partial x} = 4 = -\dot \lambda$

I've assumed that x must be the state variable, but can someone describe why it must be so?

$\frac{\partial H}{\partial y} = -2y + \lambda = 0$

Now what? Please explain as if you are talking to an idiot child, which I am.

2. Please explain as if you are talking to an idiot child, which I am.
Relax. We 're all dumb kids in this grand game called life

Now, for the functional $I=\int_0^1(4x-y^2)dt$, the argument is $x=x(t)$, so maybe it makes more sence to write $I=\int_0^1(4x-x'^2)dt$. Now, the Euler-Lagrange equation (the "derivative" of I) gives us that, at a stationary point $x$,

$\frac{\partial}{\partial t}\frac{\partial}{\partial x'}(4x-x'^2)=\frac{\partial}{\partial x}(4x-x'^2)$

or $x''(t)=-2$. Solve and use the boundary conditions.