Step 1: Find the total cost function $C(x)$

The marginal cost $MC$ is the derivative of the total cost function $C(x)$ with respect to $x$. Therefore:

$MC = \frac{dC}{dx} = 84 - 2e^{-0.01x}.$

To find the total cost function $C(x)$, integrate the marginal cost function:

$C(x) = \int MC \, dx = \int \left( 84 - 2e^{-0.01x} \right) dx.$

Solve the integral:

1. The integral of $84$ is: $\int 84 \, dx = 84x.$

2. The integral of $-2e^{-0.01x}$ is: $\int -2e^{-0.01x} \, dx = -2 \cdot \frac{e^{-0.01x}}{-0.01} = 200e^{-0.01x}.$

Combine the results:

$C(x) = 84x + 200e^{-0.01x} + C_0,$

where $C_0$ is the constant of integration. Since the fixed cost is given as $7,200$, we have:

$C(0) = 84(0) + 200e^{-0.01(0)} + C_0 = 7200.$

Simplify:

$C_0 = 7200 - 200 = 7000.$

Thus, the total cost function is:

$C(x) = 84x + 200e^{-0.01x} + 7000.$

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Step 2: Find the cost of producing 40 units

Substitute $x = 40$ into the total cost function:

$C(40) = 84(40) + 200e^{-0.01(40)} + 7000.$

Compute each term:

1. $84(40) = 3360$,
2. $e^{-0.01(40)} = e^{-0.4} \approx 0.67032$,
3. $200 \cdot 0.67032 \approx 134.064$.

Combine:

$C(40) = 3360 + 134.064 + 7000 = 10494.064.$

Rounding to the nearest dollar:

$C(40) \approx 10,494 \, \text{dollars}.$

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Final Answers:

1. Total cost function: $C(x) = 84x + 200e^{-0.01x} + 7000.$

2. Cost of producing 40 units: $C(40) \approx 10,494 \, \text{dollars}.$

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Would you like me to explain the steps further or calculate something else?

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Related Questions:

1. What is the marginal cost at $x = 40$?
2. How does the fixed cost affect the total cost function?
3. Can you find the profit function if the revenue is given?
4. How does $e^{-0.01x}$ impact the cost function over time?
5. What would be the cost for 100 units?

Tip:

Understanding the role of exponential decay $e^{-kx}$ in cost functions can help you estimate long-term cost behavior more effectively.