C H A P T E R 2 Polynomial and Rational Functions

234 Chapter 2 Polynomial and Rational Functions
67.
69.
s 16t 2 v 0 t s 0 16t 2 160t
(a)
(b)
16t 2 160t 0
16tt 10 0
t 0, t 10
It will be back on the ground in 10 seconds.
16t 2 160t > 384
16t 2 160t 384 > 0
16t 2 10t 24 > 0
t 2 10t 24 < 0
t 4t 6 < 0
L 2 50L 500 ≥ 0
4 < t < 6 seconds
By the Quadratic Formula we have:
Critical numbers: L 25 ± 55
Test: Is
Solution set:
2L 2W 100 ⇒ W 50 L 70.
LW ≥ 500
L50 L ≥ 500
L 2 50L 500 ≥ 0?
25 55 ≤ L ≤ 25 55
13.8 meters ≤ L ≤ 36.2 meters
68. s 16t2 v0t s0 16t 2 128t
(a) 16t2 128t 0
(b)
16tt 8 0
It will be back on the ground in 8 seconds.
16t 2 128t 128 < 0
Critical numbers: 4 22, 4 22
Test intervals:
, 4 22, 4 22, 4 22,
4 22,
Solution set:
16t 0 ⇒ t 0
t 8 0 ⇒ t 8
16t 2 128t < 128
L 2 220L 8000 ≥ 0
0 seconds ≤ t < 4 22 seconds
and 4 22 seconds < t ≤ 8 seconds
2L 2W 440 ⇒ W 220 L
By the Quadratic Formula we have:
Critical numbers:
Test:
Is L 2 220L 8000 ≥ 0?
Solution set:
LW ≥ 8000
L220 L ≥ 8000
71. R x75 0.0005x and C 30x 250,000
72. What is the price per unit?
P R C
When x 90,000:
75x 0.0005x 2 30x 250,000
0.0005x 2 45x 250,000
0.0005x 2 45x 250,000 ≥ 750,000
0.0005x 2 45x 1,000,000 ≥ 0
Critical numbers: x 40,000, x 50,000 (These were
obtained by using the Quadratic Formula.)
Test intervals:
By testing x-values
in each test interval in the inequality, we
see that the solution set is 40,000, 50,000 or
40,000 ≤ x ≤ 50,000. The price per unit is
p R
75 0.0005x.
x
P ≥ 750,000
0, 40,000, 40,000, 50,000, 50,000,
For x 40,000, p $55. For x 50,000, p $50.
Therefore, for 40,000 ≤ x ≤ 50,000, $50.00 ≤ p ≤ $55.00.
When x 100,000:
L 110 ± 1041
110 1041 ≤ L ≤ 110 1041
R $2,880,000 ⇒ 2,880,000
90,000
R $3,000,000 ⇒ 3,000,000
100,000
45.97 feet ≤ L ≤ 174.03 feet
$32 per unit
$30 per unit
Solution interval: $30.00 ≤ p ≤ $32.00