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?
Well, it seems like the construction company is in a bit of a pickle! Let's do some math to figure out the maximum number of days they can be late before their budget runs out.
We know that the daily fine starts at K4000 and increases by K1000 each day. So, we can set up an equation to represent this situation:
Total fines = K4000 + (K4000 + K1000) + (K4000 + K1000 + K1000) + ...
Now, we need to figure out how many days it will take for the total fines to reach K60,000. Let's see:
K4000 + (K4000 + K1000) + (K4000 + K1000 + K1000) + ... = K60,000
To make things easier, let's simplify the equation by factoring out K4000:
K4000(1 + 2 + 3 + ...) = K60,000
Now, we can use the formula for the sum of an arithmetic series:
n(n + 1)/2 = 60
Rearranging the equation, we get:
n^2 + n - 120 = 0
Let's solve this quadratic equation using the quadratic formula:
n = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values a = 1, b = 1, and c = -120, we get:
n = (-1 ± √(1^2 - 4(1)(-120))) / (2(1))
Simplifying further:
n = (-1 ± √(1 + 480)) / 2
n = (-1 ± √481) / 2
Since we're dealing with time, we can ignore the negative solution:
n = (√481 - 1) / 2
Calculating the value:
n ≈ 9.27
So, the maximum number of days the construction company can be late before running out of their budget is approximately 9 days. However, I hope they finish the project on time, because clowning around with deadlines is no laughing matter!
Given that the daily fine increases by K1000 each day, we can set up an equation to represent the total fine incurred after a certain number of days:
Total Fine = K4000 + (K4000 + K1000) + (K4000 + K1000 + K1000) + ...
We notice that the daily fine is an arithmetic sequence with a common difference of K1000. The sum of an arithmetic sequence can be calculated using the formula:
Sum = (n/2)(2a + (n-1)d),
where n is the number of terms, a is the first term, and d is the common difference.
Since we want to find the maximum number of days the company can be late, we set this sum equal to K60,000:
K60,000 = (n/2)(2(K4000) + (n-1)(K1000)).
Simplifying the equation, we get:
K60,000 = n(8000 + 1000n - 1000).
K60,000 = 1000n^2 + 7000n - 8000n.
K60,000 = 1000n^2 -1000n.
Rearranging the equation and dividing both sides by 1000:
n^2 - n - 60 = 0.
Now, we can solve this quadratic equation using factoring, completing the square, or using the quadratic formula. In this case, factoring is the most straightforward method:
(n - 6)(n + 10) = 0.
Setting each factor equal to zero and solving for n:
n - 6 = 0 or n + 10 = 0.
n = 6 or n = -10.
Since we are looking for the maximum number of days the company can be late, we consider only the positive solution:
n = 6.
Therefore, the maximum number of days the company can be late without exceeding its budget is 6 days.
To determine the maximum number of days the construction company can be late, we need to calculate how many days it will take for the total fines to reach K60,000.
Let's break down the daily fines:
- Day 1: K4000
- Day 2: K4000 + K1000 = K5000
- Day 3: K4000 + K1000 + K1000 = K6000
- Day 4: K4000 + K1000 + K1000 + K1000 = K7000
- Day 5: K4000 + K1000 + K1000 + K1000 + K1000 = K8000
- Day 6: K4000 + K1000 + K1000 + K1000 + K1000 + K1000 = K9000
We can observe that the daily fines form an arithmetic sequence with a common difference of K1000 and a first term of K4000.
The sum of an arithmetic sequence can be calculated using the formula Sn = (n/2)(2a + (n-1)d), where Sn is the sum, n is the number of terms, a is the first term, and d is the common difference.
In this case, we want to find the maximum value of n (number of days) where the sum of fines (Sn) does not exceed K60,000.
Let's substitute the given values into the formula and solve for n:
Sn = (n/2)(2a + (n-1)d)
K60,000 = (n/2)(2 * K4000 + (n-1) * K1000)
K60,000 = (n/2)(8000 + 1000n - 1000)
K120,000 = 8000n + 1000n^2 - 1000n
K1000n^2 + 7000n - K120,000 = 0
Using the quadratic formula, we can solve for n:
n = (-b ± √(b^2 - 4ac)) / (2a)
where a = 1000, b = 7000, and c = -120,000
n = (-7000 ± √(7000^2 - 4 * 1000 * -120000)) / (2 * 1000)
Simplifying the equation gives us:
n = (-7000 ± √(49000000 + 48000000)) / 2000
n = (-7000 ± √(97000000)) / 2000
n = (-7000 ± 9857.84) / 2000
We have two possible solutions:
n1 = (-7000 + 9857.84) / 2000 ≈ 1.43
n2 = (-7000 - 9857.84) / 2000 ≈ -9.43
Since the number of days cannot be negative, we take the positive solution, n ≈ 1.43.
Therefore, the maximum number of days the construction company can be late is approximately 1 day. This means that the company must complete the project within the deadline to avoid exceeding the K60,000 budget for fines.
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