# 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....

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,

let's say,

`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.

Therefore,

`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.