Respuesta :
length=x
width=y
Perimeter of a rectangle=2(length)+2(width)
Therefore:
2x+2y=200
We simplify the equation dividend both sides of this equation by 2:
x+y=100
Then:
y=100-x
Area of a rectangle: length x width
We have the next inequation:
x(100-x)<900
100x-x²<900
x²-100x+900<0
We solve this inequation
1) we solve this equation:
x²-100x+900=0
x=[100⁺₋√(10000-3600)]/2=(100⁺₋80)/2
We have two solutions in this equation:
x₁=90
x₂=10
2)With these values, we make intervals:
(-∞,10)
(10,90)
(90,∞)
3)With these intervals, we check it out if the inequation works:
(-∞,10); for example; if x=0 ⇒ 0²-100(0)+900=100>0, this interval don´t work.
(10,90);f.e: if x=11; ⇒ 11²-100(11)+900=-79<0, this interval works.
(90,∞); fe; if x=91 ⇒91²-(100)91+900=81>0; this interval don´t work.
answer: the possible lengths would be the values inside of this interval:
(10,90) ft.
width=y
Perimeter of a rectangle=2(length)+2(width)
Therefore:
2x+2y=200
We simplify the equation dividend both sides of this equation by 2:
x+y=100
Then:
y=100-x
Area of a rectangle: length x width
We have the next inequation:
x(100-x)<900
100x-x²<900
x²-100x+900<0
We solve this inequation
1) we solve this equation:
x²-100x+900=0
x=[100⁺₋√(10000-3600)]/2=(100⁺₋80)/2
We have two solutions in this equation:
x₁=90
x₂=10
2)With these values, we make intervals:
(-∞,10)
(10,90)
(90,∞)
3)With these intervals, we check it out if the inequation works:
(-∞,10); for example; if x=0 ⇒ 0²-100(0)+900=100>0, this interval don´t work.
(10,90);f.e: if x=11; ⇒ 11²-100(11)+900=-79<0, this interval works.
(90,∞); fe; if x=91 ⇒91²-(100)91+900=81>0; this interval don´t work.
answer: the possible lengths would be the values inside of this interval:
(10,90) ft.
