Answer:
Possible values of x are 50 < x < 110.
Step-by-step explanation:
Consider the perimeter of the field:
2x + 2(x + 30) < 500
2x + 2x + 60 < 500
4x < 440
x < 110.
Consider the area of the field:
x(x + 30 ) > 4000
x^2 + 30x - 4000 > 0
(x - 50)(x + 80) > 0
The critical values are x = -80 and 50.
As x is a width it must be positive so x > 50.
