Pregunta
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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 \( x=1000 \).

Ask by Griffin Schofield. in the United States
Feb 20,2025

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a. The average rate of change in revenue as \( x \) changes from 1001 to 1005 is \( 15.976 \) thousand dollars. b. The marginal revenue at \( x = 1000 \) is \( 4 \) thousand dollars.

Solución

To find the average rate of change in revenue as \( x \) changes from 1001 to 1005, we need to calculate the difference in revenue and divide it by the difference in \( x \). Given: \[ R(x) = 6x - 0.001x^2 \] a. Find the average rate of change in revenue as \( x \) changes from 1001 to 1005: \[ \text{Average Rate of Change} = \frac{R(1005) - R(1001)}{1005 - 1001} \] b. Find the marginal revenue at \( x = 1000 \): \[ \text{Marginal Revenue} = R'(x) \] Let's calculate the average rate of change and the marginal revenue. Calculate the value by following steps: - step0: Calculate: \(\left(6\times 1005-0.001\times 1005^{2}\right)-\left(6\times 1001-0.001\times 1001^{2}\right)\) - step1: Multiply the numbers: \(\left(6030-0.001\times 1005^{2}\right)-\left(6\times 1001-0.001\times 1001^{2}\right)\) - step2: Multiply the numbers: \(\left(6030-1010.025\right)-\left(6\times 1001-0.001\times 1001^{2}\right)\) - step3: Subtract the numbers: \(5019.975-\left(6\times 1001-0.001\times 1001^{2}\right)\) - step4: Multiply the numbers: \(5019.975-\left(6006-0.001\times 1001^{2}\right)\) - step5: Multiply the numbers: \(5019.975-\left(6006-1002.001\right)\) - step6: Subtract the numbers: \(5019.975-5003.999\) - step7: Subtract the numbers: \(15.976\) Calculate or simplify the expression \( 6*1005 - 0.001*1005^2 - (6*1001 - 0.001*1001^2) \). Calculate the value by following steps: - step0: Calculate: \(6\times 1005-0.001\times 1005^{2}-\left(6\times 1001-0.001\times 1001^{2}\right)\) - step1: Multiply the numbers: \(6\times 1005-0.001\times 1005^{2}-\left(6006-0.001\times 1001^{2}\right)\) - step2: Multiply the numbers: \(6\times 1005-0.001\times 1005^{2}-\left(6006-1002.001\right)\) - step3: Subtract the numbers: \(6\times 1005-0.001\times 1005^{2}-5003.999\) - step4: Multiply the numbers: \(6030-0.001\times 1005^{2}-5003.999\) - step5: Multiply the numbers: \(6030-1010.025-5003.999\) - step6: Subtract the numbers: \(15.976\) The average rate of change in revenue as \( x \) changes from 1001 to 1005 is \( 15.976 \) (or \( \frac{1997}{125} \) or \( 15\frac{122}{125} \)). Now, let's find the marginal revenue at \( x = 1000 \). Calculate the value by following steps: - step0: Calculate: \(6-2\times 0.001\times 1000\) - step1: Multiply the terms: \(6-2\) - step2: Subtract the numbers: \(4\) The marginal revenue at \( x = 1000 \) is 4.

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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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