The cost (in dollars) of producing x units of a certain commodity is C(x) = 7000 + 10x + 0.05x2.(a) Find the average rate of change of C with respect to x when the production level is changed from x = 100 to the given value. (Round your answers to the nearest cent.)(i) x = 102$ per unit(ii) x = 101$ per unit(b) Find the instantaneous rate of change of C with respect to x when x = 100. (This is called the marginal cost.)
Question
The cost (in dollars) of producing x units of a certain commodity is C(x) = 7000 + 10x + 0.05x2.(a) Find the average rate of change of C with respect to x when the production level is changed from x = 100 to the given value. (Round your answers to the nearest cent.)(i) x = 102 per unit(b) Find the instantaneous rate of change of C with respect to x when x = 100. (This is called the marginal cost.)
Solution
(a) The average rate of change of C with respect to x when the production level is changed from x = 100 to the given value can be found using the formula:
ΔC/Δx = [C(x2) - C(x1)] / (x2 - x1)
(i) For x = 102, we have:
ΔC/Δx = [C(102) - C(100)] / (102 - 100)
First, calculate the cost at x = 102 and x = 100:
C(102) = 7000 + 10102 + 0.05102^2 = 7142.04 C(100) = 7000 + 10100 + 0.05100^2 = 7100
Then, substitute these values into the formula:
ΔC/Δx = [7142.04 - 7100] / (102 - 100) = 21.02 $ per unit
(ii) For x = 101, we have:
ΔC/Δx = [C(101) - C(100)] / (101 - 100)
First, calculate the cost at x = 101:
C(101) = 7000 + 10101 + 0.05101^2 = 7110.05
Then, substitute these values into the formula:
ΔC/Δx = [7110.05 - 7100] / (101 - 100) = 10.05 $ per unit
(b) The instantaneous rate of change of C with respect to x when x = 100 can be found by taking the derivative of C(x) and evaluating it at x = 100. The derivative of C(x) is:
C'(x) = 10 + 0.1x
Evaluating this at x = 100 gives:
C'(100) = 10 + 0.1*100 = 20
So, the marginal cost when x = 100 is $20 per unit.
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