Minima or what?

Algebra Level pending

The minimum value of 2 x 2 + 2 x y + y 2 + 2 x − 3 y + 8 2x^2+2xy+y^2+2x-3y+8 for real numbers x x and y y is p p . . Then find the value of ⌊ p ⌋ \lfloor p \rfloor .

Notation : ⌊ ⋅ ⌋ \lfloor \cdot \rfloor denotes the floor function .


The answer is -1.

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Rinkon Saha by Rinkon Saha , 23, India

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1 solution

Rinkon Saha
May 15, 2016

The given expression can be written as : : 2 x 2 + 2 x y + y 2 + 2 x − 3 y + 8 = 1 2 ( 4 x 2 + 4 x y + 2 y 2 + 4 x − 6 y + 16 ) = 1 2 { ( 2 x + y + 1 ) 2 + ( y − 4 ) 2 − 1 } ≥ − 1 2 \begin{aligned} 2x^2+2xy+y^2+2x-3y+8 &= \frac{1}{2}(4x^2+4xy+2y^2+4x-6y+16)\\ &=\frac{1}{2}\{(2x+y+1)^2+(y-4)^2-1\}\geq-\frac{1}{2}\end{aligned} Therefore, least value of 2 x 2 + 2 x y + y 2 + 2 x − 3 y + 8 = − 1 2 2x^2+2xy+y^2+2x-3y+8 = -\frac{1}{2} at x = − 5 2 , y = 4 x=-\frac{5}{2}, y=4 . .

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