For what x an y is #y / ( x + 3 )^2 > 2/(x-y-3)#?

Question

1628 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-03T20:16:34

There is no pair #{x,y}# satisfying the condition.

The pairs #{x,y}# satisfying

#y / ( x + 3 )^2 > 2/(x-y-3)#

also satisfy

#y(x-y-3) > 2(x+3)^2#

and also satisfy

#y(x-y-3) -2(x+3)^2>0#

Calling now

#f(x,y) = y(x-y-3) -2(x+3)^2#

let us find the local minima/maxima: Solving the conditions

#grad f(x,y) = vec 0#

or

#{ (12 - 4 x + y = 0), (-3 + x - 2 y = 0) :} #

with solution

#{x=3, y= 0}#

for #f(x,y)# we have

#grad^2 f(x,y) =((-4, 1),(1, -2)) # with eigenvalues

#{-3 - sqrt[2], -3 + sqrt[2]}# so #f(x,y)# has a maximum which is a global maximum, at #{x=3, y= 0}#.

Evaluating #f(3,0) = 0# we see that there is no pair #{x,y}# which obeys strictly the condition.