§ Lagrange multipliers by algebra
§ Constrained optimisataion: the first stab
We want to maximize given the constraint that .
Let us arbitrarily say that is independent and is dependent on , so there
is some function such that . This lets us compute using:
Next, since is a function of alone (as is dependent on via ), the condition
guarantees a maxima for :
Solving the above condition along with to recover the value of gives us the optima.
Constrained optimisataion: equal footing
If we consider the stationary value of , we get:
to eliminate from the above equations, we rearrange:
This can be eliminated to recover the previous equation:
The langrange multipler procedure is nice since it does not break the symmetry between the two variables.