Maths

2x^2 + y^2 + 3xy - x - 1 = 0
==> (2x + y + 1)(x + y - 1) = 0

So, this equation is a pair of straight lines,

the intersection point of these two straight lines is (-2, 3).

There is no such max. or min.

2014-11-28 11:40:44 補充：
No. When x=3, y=-5, then
2x² + y² + 3xy - x - 1
= 18 + 25 - 45 - 3 - 1
= -6
So, no min or max on 2x² + y² + 3xy - x - 1.

(Straight line gets no min or max)

2x^2 + y^2 + 3xy - x - 1 = 0
(9/4)x^2+3xy+y^2 –(1/4)x^2-x-1=0
[(3/2)x+y]^2-(1/4)(x^2+4x+4)=0
[(3/2)x+y]^2-(1/4)(x+2)^2=0
[(3/2)x+y]^2-[(1/2)(x+2)]^2=0
Since both square terms are ≥0, there exist a minimum when both square terms are zero.
At the minimum point,
[(3/2)x+y]=0 and [(1/2)(x+2)] =0
x=-2, y=3

Are you asking for the max/min of

"2x² + y² + 3xy - x - 1"

rather than

"2x² + y² + 3xy - x - 1 = 0" ???

2014-11-28 01:00:31 補充：
Then the correct claim should be the minimum of 2x² + y² + 3xy - x - 1 is 0.