My problem, which I proposed to myself months ago is based on the simple optimization problem
in which you find the best path for a lifeguard to rescue a drowning victim. Obviously the
shortest path is a straight line, but the catch is he swims more slowly than he runs. I
solved the problem and then attempted with 3 different speeds (beach, water, and weeds, or
something of the sort). The difficulty didn't set in until I allowed the difficulty of
traversing a point in the plane vary continuously. How do I find an optimal path?
Unfortunately you didn't supply a figure. I take it that you have the following in mind:
At time $t=0$ the life guard is at $(0,0)$ in the $(x,y)$-plane, and the drowning swimmer is at the point $(a,b)$ in the first quadrant. The speed of the life guard is a given function $x\mapsto v(x)$ $\ (0\leq x\leq a)$ of $x$ alone.
Then one can set up the following variational problem with a constraint: Denote by $p(x)$ the slope of the life guard's orbit when his abscissa is $x$. Then $dt$ and $dx$ are related by $$dt={ds\over v(x)} ={\sqrt{1+p^2(x)}\over v(x)}\ dx\ .$$ Therefore we should minimize $$T:=\int_0^a {\sqrt{1+p^2(x)}\over v(x)}\ dx$$ under the constraint $$\int_0^a p(x)\ dx=b\ .$$ This is a standard variational problem. The constraint is dealt with using a Lagrange multiplier.
For the discontinuous version of this, meaning $v(x)=v_1$ for $0\leq x\leq c$ and $v(x)=v_2$ for $c\leq x\leq a$, see Snell's law at Wikipedia.