During the first year of production, 8 000 000 tonnes of copper were extracted from a Peruvian mine. Production has declined by 8% each year since then.a.) What was the total mass of the copper mined during the first four years of production?b.) How many years will it take for production to fall below 100 000 per year? c.) If the production trend continues, how much copper will be extracted from the mine over its lifetime?
4 years ago
Answered By Mingrui (Mark) J
1.
year 1 = 8 000 000
year 2 = (year 1) * (1- 0.08) = 8 000 000 * 0.92 = 7 360 000
year 3 = ( year 2) * (1- 0.08) = 7 360 000 * 0.92 = 6 771 200
year 4 = (year 3) * (1- 0.08) = 6 771 200 * 0.92 = 6 229 504
total = year 1 + year 2 + year 3 + year 4 = 8000000 + 7360000 + 6771200 + 6229504 = 28360704 tonnes
4 years ago
Answered By Mingrui (Mark) J
1.
year 1 = 8 000 000
year 2 = (year 1) * (1- 0.08) = 8 000 000 * 0.92 = 7 360 000
year 3 = ( year 2) * (1- 0.08) = 7 360 000 * 0.92 = 6 771 200
year 4 = (year 3) * (1- 0.08) = 6 771 200 * 0.92 = 6 229 504
total = year 1 + year 2 + year 3 + year 4 = 8000000 + 7360000 + 6771200 + 6229504 = 28360704 tonnes
4 years ago
Answered By Mingrui (Mark) J
production after x years = initial * (1-0.08)number of years
log (production after x years) = log (initial * (1-0.08)number of years) = log( initial ) + number of years * log (0.92)
number of years = ( log (production after x years) - log ( initial) ) / log (0.92) = ( log (100000) - log (8000000) ) / log(0.92) = (5-6.9031) / (-0.03621) = (-1.9031) / (-0.03621) = 52.5738 years
which rounded up to 53 years
4 years ago
Answered By Mingrui (Mark) J
check for b:
production per year = initial * (1-0.08)years = 8000000 * (0.92)52 = 104726.6 ton, which is just a bit above 100000
production per year = initial * (1-0.08)years = 8000000 * (0.92)53 = 96348.5 ton, which is just a bit below 100000
therefore 53 years would be reasonable
4 years ago
Answered By Daiwei L
a) Lets use the sum of geometric series, recall the formula as Sn = a1[(1 - rn) / (1 - r)]. We have 4 iteration for 4 years, so we have n = 4, a1 = 8 000 000 tonnes and rate of change (r) = 0.92,
plug in the formula : Sn = 8 000 000 * [(1 - (0.92)4) / (1 - (0.92))] = 28 360 704
b) Let the year of production be x, we can set up an algebraic equation.
by geomtric series to solve for the nth term, an = a1 * r(n-1)
subbing in the values, a1 = 8 000 000, r = 0.92, an = 100 000
100 000 = 8 000 000 * 0.92(n - 1)
100 000 / 8 000 000 = 0.92(n-1)
0.0125 = 0.92(n-1)
log 0.0125 = (n - 1) * log (0.92) =>by power rule(
(log(0.0125) / log (0.92)) + 1 = 53.55, we round to 54 years.
c)
To find the sum of the all of the copper mined we use a infinite sum formula = a / (1 - r)
a = 8 000 000, r = 0.92
8 000 000 / (1 - 0.92) = 8 000 000 / 0.08 = 100 000 000 tonnes over its lifetime