Example :
min $f(x,y)=x^4-y^4$ subject to $(x-5)^3-y^4≥0$
When we use the lagrangian theorem we end up with the point $(5,0)$ which is a critical point of the constraint. A necessary condition to use the lagrangian theorem was to make sure that the point we end up with is not a critical point.
I couldn't really get it. How did we end up with $(5,0)$ then? Would this mean that $(5,0)$ is not a minimiser for the function on the constraint, because the point we ended up with the lagrangian is the critical point of the constraint function?
Also another question How will I minimise $f$ on this constraint then?
The theorem says that any extremum of $f$ subject to $g = 0$ where the gradient of $g$ is not $0$ is a critical point of $f + \lambda g$. $(5,0)$ turns out to be the minimum of $f(x,y) = x^4-y^4$ subject to $g(x,y) = (x-5)^3 - y^4 = 0$, but it is not a critical point of $f(x,y) + \lambda g(x,y)$: in fact $$ \dfrac{\partial}{\partial x} (f + \lambda g)(5,0) = 500$$ This is possible because the gradient of $g$ is $0$ there.
So this is illustrating that besides the critical points of $f + \lambda g$ you have to consider points where $g=0$ and $\nabla g = 0$ as candidates for extrema.