We are given the revenue function $R(x) = 56x - 0.020x^2$, where $R(x)$ is the revenue (in dollars) from the sale of $x$ car seats, and $0 \leq x \leq 2800$.

Part (A): Find the average change in revenue if production is changed from 1,000 car seats to 1,050 car seats.

The average change in revenue between two points $x_1$ and $x_2$ is given by the formula:

$\text{Average change in revenue} = \frac{R(x_2) - R(x_1)}{x_2 - x_1}$

Here, $x_1 = 1000$ and $x_2 = 1050$, so we need to calculate $R(1000)$ and $R(1050)$.

• $R(1000) = 56(1000) - 0.020(1000)^2 = 56000 - 0.020(1000000) = 56000 - 20000 = 36000$
• $R(1050) = 56(1050) - 0.020(1050)^2 = 58800 - 0.020(1102500) = 58800 - 22050 = 36750$

Now, calculate the average change:

$\text{Average change in revenue} = \frac{R(1050) - R(1000)}{1050 - 1000} = \frac{36750 - 36000}{50} = \frac{750}{50} = 15$

So, the average change in revenue when production is increased from 1,000 to 1,050 car seats is $15 per car seat.

Part (B): Use the four-step process to find $R'(x)$.

The four-step process involves:

1. Identify the function: The given function is $R(x) = 56x - 0.020x^2$.

2. Find the derivative: We need to differentiate the function with respect to $x$.

$R'(x) = \frac{d}{dx} (56x - 0.020x^2)$

Using standard differentiation rules:

• The derivative of $56x$ is 56.
• The derivative of $-0.020x^2$ is $-0.040x$ (using the power rule).

So,

$R'(x) = 56 - 0.040x$

Thus, the derivative $R'(x) = 56 - 0.040x$.

3. Interpret the derivative: The derivative $R'(x)$ represents the instantaneous rate of change of revenue with respect to the number of car seats produced.

4. Express the final answer: The derivative of the revenue function is:

$R'(x) = 56 - 0.040x$

Part (C): Find the revenue and the instantaneous rate of change of revenue at a production level of 1,000 car seats, and interpret the results.

First, let's find the revenue at $x = 1000$:

$R(1000) = 56(1000) - 0.020(1000)^2 = 56000 - 20000 = 36000$

So, the revenue at 1,000 car seats is $36,000.

Now, let's find the instantaneous rate of change of revenue at $x = 1000$ by calculating $R'(1000)$:

$R'(1000) = 56 - 0.040(1000) = 56 - 40 = 16$

So, the instantaneous rate of change of revenue at a production level of 1,000 car seats is $16 per car seat.

Interpretation:

• The revenue at a production level of 1,000 car seats is $36,000.
• The instantaneous rate of change of revenue at this point is $16 per additional car seat. This means that if the company produces one more car seat (going from 1,000 to 1,001 car seats), the revenue will increase by approximately $16.

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Follow-up Questions:

1. How would the revenue change if the production level were increased from 1,500 to 1,550 car seats?
2. How does the instantaneous rate of change of revenue at a production level of 2,000 compare to that at 1,000?
3. What is the maximum revenue that can be obtained from this production function?
4. How would the average change in revenue behave if the production were increased from 500 to 550 car seats?
5. What does the negative coefficient in front of $x^2$ in the revenue function suggest about the nature of the revenue as production increases?

Tip:

When calculating the instantaneous rate of change (the derivative), remember that it gives you the slope of the tangent line at any point on the graph of the function. This slope represents how the output (revenue, in this case) changes for a small change in the input (the number of car seats).