An object is free to move within the set $S=$ { $(x, y) \in R^{2} | x+y \leq \cos (y-x) $ }.
It tries to come as close as possible to the point $\mathrm{P}=(1,1) > .$ Where should it go? In order to solve the problem, sketch the set $\mathrm{S}$.
[Hint: A linear co-ordinate transform might help you to understand the situation. Although the set $\mathrm{S}$ is NOT convex, there is only one local minimum, which is also the global one we are looking for.]
Mind: This problem is not convex. Still...
- Sketch a qualitatively correct picture of the set $\mathrm{S}$ and the point $\mathrm{P}$
- Explain why the constraint should be binding.
- Write the problem in standard form with Lagrange multiplier $\lambda,$ and show that there is only a single solution which is then the desired minimum.
I wrote the optimization as
min $(x-1)^2+(y-1)^2$
s.t. $x+y-cos(y-x) \leq 0$
and used KKT condition to solve this problem, but found the conditions are hard to compute because of the existence of $cos(y-x)$. (later I checked that KKT can be applied only when $(x,y)$ is from a convex set, so here is not the case.)
I could not figure out how to do the co-ordinate transformation and sketch the set $S$ either :( Could anyone shed some light on this? Thanks!
Hint.
Making the change in coordinates
$$ u = y+x\\ v = y-x $$
the problem becomes
$$ \min \frac 12\left((u-2)^2+v^2\right)\ \ \ \text{s. t. }\ \ \ \ u-\cos v \le 0 $$
This problem has a symmetry regarding the axis $v = 0$