The rate of expenditure for maintenance of a particular machine is given by M'(x)=(sqrt (x^2+8x))*(2x+8), where x is time measured in years.
Aggregate maintenance costs after the 4th year $540.
a)Find aggregate maintenance function.
b) How many years must pass before the aggregate maintenance costs reach $2000?
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The rate of expenditure in year x is, M'(x)
`M'(x) = (sqrt(x^2+8x))*(2x+8)`
If we integrate this we get the aggregate maintenance function over the years, let's name it as M(x)
`M(x) = int(sqrt(x^2+8x))*(2x+8) dx`
this can be easily doen through a simpel subsitution,
`u = x^2+8` then, `du = (2x+8)dx`
Therefore the integral changes
`M(x) = intsqrtu*du`
`M(x) = u^(3/2)/(3/2)+C`
C is the constant, we have to find it through given data.
`M(x) = (2(x^2+8)^(3/2))/3+C`
now we are given one data that, after 4 years, the aggregate maintenance cost is 540.
`540 = (2(4^2+8)^(3/2))/3+C`
`540 = 78.3837 + C`
`C = 461.6163`
a)therefore the aggregate total function is,
`M(x) = (2(x^2+8)^(3/2))/3+461.6163`
b) To find the year in which aggregate cost are equal to 2000, you have to solve for x
`2000 = (2(x^2+8)^(3/2))/3+461.6163`
`(2(x^2+8)^(3/2))/3 = 1538.3837`
`(x^2+8)^(3/2) = 2307.57555`
`x^2+8 = 174.624`
`x = 12.908`
Therefore the 12.9 (or 13 years) should pass for aggregate costs to reach 2000.
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