The cost "C" of operating a concrete-cutting machine is related to the number of minutes "n" the machine is run by the function: C(n)=2.2n⋀2-66n+655. For what number of minutes is the cost of running the machine a minimum? What is the minimum cost?
dC/dn = 4.4n - 66 = 0 for a min of C
4.4n = 66
n = 66/4.4 = 15
cost(15) = ..... put 15 back into original equation
To find the number of minutes at which the cost of running the machine is a minimum, we need to find the value of "n" for which the derivative of the cost function, C'(n), is equal to 0. After finding this value, we can substitute it back into the original cost function to find the minimum cost.
Step 1: Take the derivative of the cost function with respect to "n".
C(n) = 2.2n^2 - 66n + 655
We'll take the derivative term by term:
C'(n) = d(2.2n^2)/dn - d(66n)/dn + d(655)/dn
Differentiating each term:
C'(n) = 4.4n - 66 + 0
C'(n) = 4.4n - 66
Step 2: Set C'(n) = 0 and solve for "n".
4.4n - 66 = 0
4.4n = 66
n = 66 / 4.4
n = 15
Step 3: Substitute n = 15 back into the original cost function to find the minimum cost.
C(n) = 2.2n^2 - 66n + 655
C(15) = 2.2(15)^2 - 66(15) + 655
C(15) = 2.2(225) - 990 + 655
C(15) = 495 - 990 + 655
C(15) = 160
Therefore, the cost of running the machine is at its minimum when it is operated for 15 minutes. The minimum cost is $160.