a coffee company has found that the marginal cost, in dollars per pound, of the coffee it roasts is…

a coffee company has found that the marginal cost, in dollars per pound, of the coffee it roasts is represented by the function below, where x is the number of pounds of coffee roasted. find the total cost of roasting 200 lb of coffee, disregarding any fixed costs.\nc(x)= - 0.010x + 8.50, for x ≤ 500\nthe total cost is $□\n(round to the nearest cent as needed.)

a coffee company has found that the marginal cost, in dollars per pound, of the coffee it roasts is represented by the function below, where x is the number of pounds of coffee roasted. find the total cost of roasting 200 lb of coffee, disregarding any fixed costs.\nc(x)= - 0.010x + 8.50, for x ≤ 500\nthe total cost is $□\n(round to the nearest cent as needed.)

Answer

Explanation:

Step1: Integrate the marginal cost function

The total cost (C(x)) (ignoring fixed costs) is the integral of the marginal cost function (C^{\prime}(x)). We know that (\int C^{\prime}(x)dx=\int(- 0.010x + 8.50)dx). Using the power rule (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C) ((n\neq - 1)), we have (\int(-0.010x + 8.50)dx=-0.010\times\frac{x^{2}}{2}+8.50x+C). Since we are ignoring fixed costs ((C = 0)), (C(x)=-0.005x^{2}+8.50x).

Step2: Evaluate the definite integral from (x = 0) to (x = 280)

We use the fundamental theorem of calculus (\int_{a}^{b}C^{\prime}(x)dx=C(b)-C(a)). Here, (a = 0) and (b = 280). (C(280)-C(0)=\left[-0.005\times(280)^{2}+8.50\times280\right]-\left[-0.005\times(0)^{2}+8.50\times0\right]). First, calculate (-0.005\times(280)^{2}=-0.005\times78400=-392). Second, calculate (8.50\times280 = 2380). Then (C(280)-C(0)=-392 + 2380=1988).

Answer:

(1988)