+1533 Find the minimum value of the expression $x^2+y^2+2x-4y+8+10x-12y$ for real $x$ and $y$.
+129850 x^2 + y^2 + 2x - 4y + 8 + 10x -12y
x^2 + 12x + y^2 - 16y + 8
Take the derivative with respect to x and set to 0
2x + 12 = 0 → x + 6 = 0 → x = -6
Take the derivative with respect to y and set to 0
2y - 16 = 0 → y - 8 = 0 → y = 8
The minimum occurs at (-6,8)
And the minimum is
(-6)^2 + 12 (-6) + (8)^2 - 16(8) + 8 = -92
