The revenue (in thousands of dollars) from producing \( x \) units of an item is modeled by \( R(x)=6 x-0.001 x^{2} \). a. Find the average rate of change in revenue as \( x \) changes from 1001 to 1005 . b. Find the marginal revenue at \( x=1000 \).
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Beyond the Answer
To find the average rate of change in revenue as \( x \) changes from 1001 to 1005, we use the formula for the average rate of change: \[ \text{Average Rate of Change} = \frac{R(1005) - R(1001)}{1005 - 1001} \] First, calculate \( R(1001) \) and \( R(1005) \): \[ R(1001) = 6(1001) - 0.001(1001)^2 = 6006 - 1002.001 = 5003.999 \] \[ R(1005) = 6(1005) - 0.001(1005)^2 = 6030 - 1002.025 = 5027.975 \] Now, substituting these values back into the average rate of change formula: \[ \text{Average Rate of Change} = \frac{5027.975 - 5003.999}{4} = \frac{23.976}{4} \approx 5.994 \] So, the average rate of change in revenue as \( x \) changes from 1001 to 1005 is approximately $5.994 thousand per unit. --- To find the marginal revenue at \( x=1000 \), we calculate the derivative \( R'(x) \): \[ R'(x) = \frac{d}{dx}(6x - 0.001x^2) = 6 - 0.002x \] Now substituting \( x = 1000 \): \[ R'(1000) = 6 - 0.002(1000) = 6 - 2 = 4 \] Therefore, the marginal revenue at \( x=1000 \) is $4 thousand per unit.
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