Consider the problem: $$\min_{x, y} \{(x-1)^2 + (y-2)^2\}$$ subject to $$ \begin{cases} 2x + y \leq 4 \\ 3x + 10y \leq 20 \\ x, y \geq 0 \end{cases} $$ I have to solve this.
Attempts
So first of all I notice that without the constraints the min would lie at $(0, 0)$ since the function is a sum of two squares.
Anyway, I drew the constraints, which are two lines, and the lie in the first quadrant, since $x, y \geq 0$.
By the drawing I can conclude that only the first constraint is active, since the region created by the first constraint is contained in the region created by the second constraint, hence the feasible region is the first one (satisfying both constraints).
Now, since the two constraints are linear, I understood that max/min must lie on the boundary (if I'm wrong please correct me, and also I wouldn't mind some clarification about, either ways).
Now by Lagrange function I wrote:
$$L = (x-1)^2 + (y-2)^2 - \lambda(x + 2y - 4) - \mu(3x + 10y - 20)$$
Whence the Lagrange equations read
$$ \begin{cases} 2(x-1) - \lambda - 3 \mu = 0 \\ 2(y-2) - 2\lambda - 10\mu = 0 \\ x + 2y - 4 = 0 \\ 3x + 10y - 20 = 0 x \geq 0\\ y \geq 0 \end{cases} $$
Yet now I am stuck because I do not know how to exactly proceed.
I solved for the parameters:
$$\lambda = \dfrac{1}{2}(10x - 3y - 4) \qquad \qquad \mu = \frac{1}{2}y - x$$
Or else for $y = \lambda + 5\mu$ in terms of the multipliers.
Since only the first constraint is active, shall I forget about the second one? Also, how to proceed in order to find a numerical solution?
Thank you
According to the discussion in the comments, the minimum point is along the boundary which is made of four segments. As regards the constraint $3x + 10y=20$, you should consider the following system plus the extreme points: $$\begin{cases} 2(x-1) - 3 \mu = 0 \\ 2(y-2) - 10\mu = 0 \\ 3x + 10y - 20 = 0\\ 2x + y \leq 4, x \geq 0, y \geq 0 \end{cases}$$ Can you take it from here?