Hi
Is there a method to find out all ordered pairs for which the inequality given by,
2m+2n-mn is always greater than 3?
that is 2m + 2n - mn >= 3
Thanks folks,
Okay, I had a sign error of all things!
The simplest way to handle a complicated inequality is to first solve the associated equation.
That is, look at 2m+ 2n- mn= 3 or, preferably, xy- 2x- 2y+ 3= 0.
If we let x= x'- a and y= y'- b then xy- 2x- 2y+ 3= x'y'- ax'- ay'+ab- 2x'+ 2a- 2y'+ 2b+ 3= x'y'- (a+2)x'- (b+2)y'+ (3+ 2a+ 2b+ ab)= 0. In particular, taking a= b= -2 reduces that to x'y'+ 3- 4= x'y' - 1= 0 or x'y'= 1. That is a hyperbola with asymptotes parallel to the x and y axes and center at (0, 0) in the x'y'-system or (2, 2) in the xy-system.
The point is that the points where that is equal to 3 separate "< 3" from "> 3". Since (2, 2) is "between" the two branches and 2(2)+ 2(2)- (2)(2)= 4> 3 it follows that 2x+ 2y- xy> 3 for all (x, y) between the two branches of the parabola. On the other hand, (8, 8) is outside one branch of the hyperbola and 2(8)+ 2(8)- (8)(8)= 32- 64= -32< 3 and (-8, -8) is outside the other branch and 2(-8)- 2(-8)- (-8)(-8)= -32- 64= -96< 3 so points between the two branches of the hyperbola are the only (x, y) that satisfy the inequality.
Thanks for the reply HALLSOFIVY...
that certainly helped a lot.
I was wondering if there is another approach which can involve principles of differentiation if it can help give an exact solution set for this problem.
I'm not sure of its existence, though.
Thanks