Conditional Extreme. Find a point in $\mathbb{R^2}$ that has the smallest sum of squared distances from the lines $x=0,y=0, x-y+1=0.$

Question

I can find the main function, but I do not know the condition, to set up the Lagrange equation. Can anyone see, what condition the point has to satisfy here?(So as to apply the Lagrange multiplier method, which I must use.) I know the point is in a triangle in the positive quadrant, or so it seems.

Answer 1

$$\mbox{Let}\:\:\:\mbox{d}(x,y) = x^2 + y^2 + \frac{(x - y + 1)^2}{2}.$$

$$\mbox{Let}\:\:\:\mbox{d}_x = 2x + (x-y+1)=3x-y+1 =0$$

$$\mbox{and}\:\:\:\mbox{d}_y = 2y-x+y-1=3y-x-1=0.$$

$$\Downarrow$$

$$\bigg(-\frac{1}{4},\frac{1}{4}\bigg) \:\:\: \mbox{is a critical point of d.}$$

No need for checking the boundaries.

Therefore, this critical point is the point we are looking for.