Questions: b) Find the instantaneous rate of change function for revenue. c) Find the revenue and the instantaneous rate of change of revenue at the production level of 1,000 car seats and interpret both results. The price-demand function (in thousands of dollars) when producing x units of an item is p=10-0.002x. a) Find the average rate of change of revenue when production is increased from 1000 to 1001 units. Interpret the result. b) Find marginal revenue when 1000 units are produced. Interpret the result.

Solution Steps

For question 10: a) To find the average rate of change of revenue when production is increased from 1000 to 1001 units, we need to calculate the revenue at both production levels and then find the difference divided by the change in units. b) To find the marginal revenue when 1000 units are produced, we need to find the derivative of the revenue function with respect to \( x \) and then evaluate it at \( x = 1000 \).

Given the price-demand function \( p = 10 - 0.002x \), the revenue function is \( R(x) = x \cdot p = x(10 - 0.002x) \).

For \( x = 1000 \): \[ R(1000) = 1000 \cdot (10 - 0.002 \cdot 1000) = 1000 \cdot 8 = 8000.0 \]

For \( x = 1001 \): \[ R(1001) = 1001 \cdot (10 - 0.002 \cdot 1001) = 1001 \cdot 7.998 = 8005.998 \]

The average rate of change of revenue from 1000 to 1001 units is given by: \[ \frac{R(1001) - R(1000)}{1001 - 1000} = \frac{8005.998 - 8000.0}{1} = 5.998 \]

The marginal revenue is the derivative of the revenue function \( R(x) \) with respect to \( x \): \[ R(x) = x(10 - 0.002x) \] \[ R'(x) = \frac{d}{dx} [x(10 - 0.002x)] = 10 - 0.004x \]

Evaluating the marginal revenue at \( x = 1000 \): \[ R'(1000) = 10 - 0.004 \cdot 1000 = 6.000 \]

Final Answer