Find the minimum value of x2+2xy+3y2−6x−2y, over all real numbers x and y.
Try to rewrite the expression as the sum of squares plus a number.
Is the answer -11? Inspiration: https://www.quora.com/If-x-and-y-are-real-numbers-then-what-is-the-minimum-value-of-x-2-+4xy+6y-2-4y+4
Don't take my word for it, though.
I think it's x=-4, y=1.
x2+2xy+3y2−6x−2y=(x+y)2+2y2−2y−6x=(x+y)2+2y(y−1)−6xMinimum occurs when x = -y.=2y2−2y+6y=2(y2+2y)=2((y+1)2−1)=2(y+1)2−2Minimum value=−2.
