Question

The least value of $2x^2 + y^2 + 2xy + 2x - 3y + 8$ for real numbers $x$ and y is

• A. 2
• B. 8
• C. 3
• D. -0.5

Answer 1

Answer: (D)

-0.5

Answer 2

$2 x^{2}+y^{2}+2 x y+2 x-3 y+8$
$=\frac{1}{2}\left[4 x^{2}+2 y^{2}+4 x y+4 x-6 y+16\right]$
$=\frac{1}{2}\left[\left(y^{2}-8 y\right)+\left(4 x^{2}+y^{2}+4 x y+4 x+2 y\right)+16\right]$ $=\frac{1}{2}\left[(y-4)^{2}+(2 x+y+1)^{2}-1\right]$
So, least value will be $-\frac{1}{2}$ at $y=4$
and $x=\frac{-5}{2}$.