Asked by Jamie-anne
A construction company will be fined for each day it is late completing its current project.The daily fine will be K4000 for the first day and will increase by K1000 each day.Based on its budget, the company can only afford K60,000 in total fines. What is the maximum number of days it can be late?
Answers
Answered by
Reiny
you have an AS
sum = 4000 + 5000 + 6000 + ..
sum(n) = (n/2)(8000 + (n-1)(1000) ) = 60000
n(8000 + 1000n - 1000) = 120000
1000n(7 + n) = 120000
n(n+7) = 120
n^2 + 7n - 120 = 0
(n - 8)(n + 15) = 0
n = 8 or n = -15, ignore the negative
They can be late 8 days
sum = 4000 + 5000 + 6000 + ..
sum(n) = (n/2)(8000 + (n-1)(1000) ) = 60000
n(8000 + 1000n - 1000) = 120000
1000n(7 + n) = 120000
n(n+7) = 120
n^2 + 7n - 120 = 0
(n - 8)(n + 15) = 0
n = 8 or n = -15, ignore the negative
They can be late 8 days
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