A factory installs new machinery that will save 1250−38x dollars per year, where x is the number of years since installation. However, the cost of maintaining the new machinery is 81x dollars per year. Find the accumulated savings that will occur before the machinery should be replaced.
To find the accumulated savings, we need to calculate the savings each year and sum them up until the machinery needs to be replaced.
The savings per year is given by the difference between the amount saved and the cost of maintenance: 1250 - 38x - 81x.
To determine when the machinery should be replaced, we set the savings per year equal to zero and solve for x: 1250 - 38x - 81x = 0.
Combining like terms, we get: -119x + 1250 = 0.
To solve for x, we can isolate x by subtracting 1250 from both sides: -119x = -1250.
Dividing by -119, we find: x = -1250 / -119.
Simplifying, we get: x ≈ 10.5042.
Since x represents the number of years since installation, we round up to the nearest whole number. Thus, the machinery should be replaced after 11 years.
To calculate the accumulated savings, we sum the savings per year for the first 11 years: 1250 - 38(1) - 81(1) + 1250 - 38(2) - 81(2) + ... + 1250 - 38(11) - 81(11).
Simplifying, we have: (1250 - 38(1) - 81(1)) + (1250 - 38(2) - 81(2)) + ... + (1250 - 38(11) - 81(11)).
Using the formula for the sum of an arithmetic series, with a = 1250 - 38 - 81 = 1131, d1 = -38 - 81 = -119, and n = 11, we can calculate the sum:
sum = (n/2)(2a + (n-1)d) = (11/2)(2(1131) + (11-1)(-119)).
Simplifying, we find:
sum = (11/2)(2262 + 10(-119)) = (11/2)(2262 - 1190) = (11/2)(1072) = 11(536) = 5896.
Therefore, the accumulated savings that will occur before the machinery needs to be replaced is $5896.
To find the accumulated savings that will occur before the machinery should be replaced, we need to determine the number of years it will take for the accumulated savings to equal the cost of replacing the machinery.
First, let's set up an equation to represent the accumulated savings:
Accumulated savings = Savings per year - Maintenance cost per year
Substituting the given values into the equation:
Accumulated savings = (1250 - 38x) - (81x)
Now, let's solve for x:
Accumulated savings = 1250 - 38x - 81x
Accumulated savings = 1250 - 119x
To find the number of years it will take for the accumulated savings to equal the cost of replacing the machinery, we need to set the equation equal to 0:
1250 - 119x = 0
Now, solve for x:
119x = 1250
x = 1250 / 119
x ≈ 10.5
Therefore, it will take approximately 10.5 years for the accumulated savings to equal the cost of replacing the machinery.
To find the accumulated savings, substitute x = 10.5 back into the equation:
Accumulated savings = 1250 - 119(10.5)
Accumulated savings ≈ 1250 - 1254.5
Accumulated savings ≈ -4.5
The negative value means that the accumulated savings is negative, indicating that the cost of replacing the machinery will exceed the savings.
When should it be replaced?
when maintenace costs are >= savings
81x=1250-38x
119x=1250
or time=1250/119=10.5 years
Accumulated savings=sum(savings-cost) from n=1 to 10
= Sum1,10 (1250-38x-81x)
Add the series of answers up for all the years, and that is the total net savings over the course of purchasing the new machinery
check my work