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

202 Chapter 2 Polynomial and Rational Functions
105.
106.
108. (a)
76x 3 4830x 2 2,820,000 0
P 76x 3 4830x 2 320,000, 0 ≤ x ≤ 60
2,500,000 76x 3 4830x 2 320,000
The zeros of this equation are x 46.1, x 38.4, and x 21.0. Since 0 ≤ x ≤ 60, we
disregard x 21.0. The smaller remaining solution is x 38.4. The advertising expense is $384,000.
The zeros of this equation are x 18.0, x 31.5, and
x 42.0. Because 0 ≤ x ≤ 50, disregard x 18.02.
The smaller remaining solution is x 31.5, or an
advertising expense of $315,000.
(b)
P 45x 3 2500x 2 275,000 107. (a) Current bin: V 2 3 4 24 cubic feet
800,000 45x 3 2500x 2 275,000
A 250 x160 x 1.5160250 (c)
60,000 x 2 410x 40,000
x 410 ± 4102 4120,000
21
x must be positive, so
x
0 45x 3 2500x 2 1,075,000
0 9x 3 500x 2 215,000
0 x 2 410x 20,000
410 ± 248,100
2
410 248,100
2
44.05.
60,000
The new length is 250 44.05 294.05 ft and the
new width is 160 44.05 204.05 ft, so the new
dimensions are 204.05 ft 294.05 ft.
109.
C is minimum when 3x3 40x2 C 100
2400x 36000 0.
200 x
, x ≥ 1
x2 x 30
The only real zero is x 40 or 4000 units.
(b)
A 250 2x160 x 60,000
x 570 ± 5702 4220,000
22
x must be positive, so
x
New bin: V 524 120 cubic feet
x 3 9x 2 26x 24 120
x 3 9x 2 26x 96 0
The only real zero of this polynomial is x 2. All the
dimensions should be increased by 2 feet, so the new
bin will have dimensions of 4 feet by 5 feet by 6 feet.
2x 2 570x 20,000 0
570 484,900
4
V 2 x3 x4 x 120
31.6.
The new length is 250 231.6 313.2 ft and the new
width is 160 31.6 191.6 ft, so the new dimensions
are 191.6 ft 313.2 ft.
110. h(t 16t2 48t 6
Let h 64 and solve for t.
64 16t 2 48t 6
16t 2 48t 58 0
48 ± i1408
By the Quadratic Formula we have t .
32
Since the equation yields only imaginary zeros, it is not
possible for the ball to have reached a height of 64 feet.