TITLE: Minimize $x^2+y^2$, subject to... (optimal points, KKT conditions, dual theories)
QUESTION [2 upvotes]: I am new to this. I am self learning to get ahead of my next years course and came across this question. I thought it would be a good question to look at due to it touching an many different aspects of optimization.

Minimize $x^2+y^2$
subject to $(x−1)^2 +(y−1)^2\leq 1$,
subject to
$(x−1)^2 +(y+1)^2\leq 1$, where $(x, y)\in\Bbb R^2$.
(i) What are the set of
feasible points for this problem? Using this, find the optimal point.
(ii) Write down the KKT conditions; are these conditions satisfied at
the optimal point? Describe why.
(iii) Write the Lagrange dual
problem, and find the optimal solution to the dual problem. Is this
optimum solution attained? Does the strong duality theorem for convex
programming apply to this scenario?

I have researched the basics of this course, and am trying to challenge myself with this question. Any help would be greatly appreciated.

REPLY [0 votes]: $d^2=x^2+y^2$ is the square of distance from $(0,0)$ to $(x,y)$.
$$\\$$
1. When $(x,y)$ is within $\space (x−1)^2+(y−1)^2≤1$,
then $d^2$ is minimum when $(x,y)=(1-\frac{\sqrt2}{2},1-\frac{\sqrt2}{2})$. 
(Imagine which point on the circle is closest from the origin.)
$\therefore d^2 \geq \left(1-\frac{\sqrt2}{2}\right)^2=\frac{3}{2}-\sqrt2$
$$\\$$
2. When $(x,y)$ is within $\space (x−1)^2+(y+1)^2≤1$,
then the minimum $d^2$ is minimum when $(x,y)=(1-\frac{\sqrt2}{2},-1+\frac{\sqrt2}{2})$. And the minimum $d^2$ is same to the above case.
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