Respuesta :
Given:
The core diameter is
D = 4 cm = 400 mm
Each ticket has length of
L = 3 in = 3*254 = 762 mm,. and thickness of
t = 0.28 mm.
In order to have 1000 tickets, the unrolled length should be
1000 * 762 = 762,000 mm
1st wrap:
diameter = D
length = πD
2nd wrap:
diameter = D + 2t
length = π(d + 2t)
3rd wrap:
diameter = D + 2(2t)
length = π[D + 2(2t)]
...
n-th wrap:
diameter = D + (n-1)*(2t)
length = π[D + (n-1)*(2t)]
The length forms an arithmetic sequence, with
a₁ = πD = 1256.6 mm (first term)
d = 2πt = 1.7593 mm
The n-th term is
1256.6 + 1.7593(n-1) = 1254.8 + 1.7593n
The total length of n wraps is
(n/2)*(1256.6 + 1254.8 + 1.7593n)
= 1255.7n + 0.8797n²
The total length should be equal to 762,000.
Therefore
0.8797n² + 1255.7n - 762000 = 0
Solve with the quadratic formula.
n = (1/1.7594)*[-1255.7 +/- √(1255.7² + 2.6812 x 10⁶ )] = 459.14 or -1886.56
Reject negative length, so that
n = 459.14, rroundto the larger value of 460.
The number of tickets will be 1000 if the 460-th wrap (outer wrap) has a diameter of
1254.8 + 1.7593*460 = 2064.1 mm, or
2064.1/254 = 8.13 in
Answer:
The diameter of each roll should be about 8.1 in (or 2064 mm)
The core diameter is
D = 4 cm = 400 mm
Each ticket has length of
L = 3 in = 3*254 = 762 mm,. and thickness of
t = 0.28 mm.
In order to have 1000 tickets, the unrolled length should be
1000 * 762 = 762,000 mm
1st wrap:
diameter = D
length = πD
2nd wrap:
diameter = D + 2t
length = π(d + 2t)
3rd wrap:
diameter = D + 2(2t)
length = π[D + 2(2t)]
...
n-th wrap:
diameter = D + (n-1)*(2t)
length = π[D + (n-1)*(2t)]
The length forms an arithmetic sequence, with
a₁ = πD = 1256.6 mm (first term)
d = 2πt = 1.7593 mm
The n-th term is
1256.6 + 1.7593(n-1) = 1254.8 + 1.7593n
The total length of n wraps is
(n/2)*(1256.6 + 1254.8 + 1.7593n)
= 1255.7n + 0.8797n²
The total length should be equal to 762,000.
Therefore
0.8797n² + 1255.7n - 762000 = 0
Solve with the quadratic formula.
n = (1/1.7594)*[-1255.7 +/- √(1255.7² + 2.6812 x 10⁶ )] = 459.14 or -1886.56
Reject negative length, so that
n = 459.14, rroundto the larger value of 460.
The number of tickets will be 1000 if the 460-th wrap (outer wrap) has a diameter of
1254.8 + 1.7593*460 = 2064.1 mm, or
2064.1/254 = 8.13 in
Answer:
The diameter of each roll should be about 8.1 in (or 2064 mm)
