A construction company will be fined each day of delay in construction for a commercial building. The penalty will be Php5,000 for the first day and will increase by Php1000 for each following day. Based on its budget, the company can afford to pay a maximum of Php50,000 for the fines. Find the maximum number of days by which the completion of work can be delayed.
since this is just an AP with
a = 5000
d = 1000
you want to find n such that
n/2 (2*5000 + 1000(n-1)) <= 50000
or, getting rid of all those zeroes,
n/2 (10+n-1) <= 50
n <= 6
check:
6 days of fines: 5 + 6+7+8+9+10 = 45
7 days = 45 + 11 = 56
I need the answer
To find the maximum number of days by which the completion of work can be delayed, we need to determine how many days the construction company can afford to pay the fines without exceeding their budget of Php50,000.
Let's assume the number of days of delay is represented by 'n'.
The penalty for the first day of delay is Php5,000.
For each subsequent day of delay, the penalty increases by Php1,000.
So, the penalty for the second day of delay is Php5,000 + Php1,000 = Php6,000.
For the third day of delay, the penalty is Php5,000 + Php1,000 + Php1,000 = Php7,000.
We can see that the penalty increases by Php1,000 for each additional day of delay.
To solve this problem, we can set up an equation to represent the total amount of fines paid for 'n' days of delay:
Total Fines = Penalty for the first day + Penalty for the second day + ... + Penalty for the nth day
Using the arithmetic series formula, we can write:
Total Fines = n/2 * (2 * First Penalty + (n-1) * Penalty Increase)
In this case:
First Penalty = Php5,000
Penalty Increase = Php1,000
So, the equation becomes:
Total Fines = n/2 * (2 * 5,000 + (n-1) * 1,000)
We want to find the maximum number of days 'n' that the company can afford, which means the total fines should not exceed Php50,000.
Setting up the inequality:
n/2 * (2 * 5,000 + (n-1) * 1,000) <= 50,000
Simplifying the equation:
n * (10,000 + 1,000n - 1,000) <= 100,000
n * (10,000 + 1,000n - 1,000) <= 100,000
n * (1,000n + 9,000) <= 100,000
1,000n^2 + 9,000n - 100,000 <= 0
Using the quadratic formula, we can solve for 'n':
n = (-b ± √(b^2 - 4ac)) / 2a
For the quadratic equation:
a = 1,000
b = 9,000
c = -100,000
Calculating the discriminant (b^2 - 4ac):
Discriminant = 9,000^2 - 4 * 1,000 * -100,000
Discriminant = 81,000,000 - (-400,000,000)
Discriminant = 481,000,000
Since the discriminant is positive, we have two real solutions for 'n'.
Applying the quadratic formula:
n = (-9,000 ± √(481,000,000)) / (2 * 1,000)
n = (-9,000 ± √481,000,000) / 2,000
Calculating the square root:
√481,000,000 ≈ 21,925.8
Substituting the values of 'n':
n = (-9,000 + 21,925.8) / 2,000 ≈ 6.96 (approximately)
n = (-9,000 - 21,925.8) / 2,000 ≈ -15.46 (approximately)
Since we are dealing with the number of days, 'n' cannot be negative. Therefore, we take the positive value:
n ≈ 6.96
Therefore, the maximum number of days by which the completion of work can be delayed is approximately 6 days.