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

1 Answer
Cesareo R.
Jul 11, 2016

See below

Explanation:

Compacting the inequality we have

#-(y (18 - x + 12 x^3 + 2 x^6 + 3 y + y^2))/((3 + x^3)^2 (x - 3 y - y^2) )>0# or
#-(y (18 - x + 12 x^3 + 2 x^6 + 3 y + y^2))/(x - 3 y - y^2)>0#
because

#(3 + x^3)^2 ge 0 forall x in RR#

The feasible set frontier is composed of

#f_1(x,y)=-y = 0#
#f_2(x,y)= 18 - x + 12 x^3 + 2 x^6 + 3 y + y^2=0#

and

#f_3(x,y) =x - 3 y - y^2=0#

The feasible region interior is the set of points #{x,y}# obeying

#f_1(x,y)f_2(x,y) > 0 and f_3(x,y)>0#

or

#f_1(x,y)f_2(x,y) < 0 and f_3(x,y)<0#

This set is shown in the attached figure in light blue

enter image source here

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