If the total revenue function for a computer is 
R(x) = 1800x − 23x2 − x3,
 find the level of sales, x, that maximizes revenue and find the maximum revenue in dollars.
x	 = 		 	computers
R(x)	 = 	$	 	


To find the level of sales $ x $ that maximizes revenue and to determine the maximum revenue, we need to follow these steps:

### 1. Find the derivative of the revenue function
The given revenue function is:
\[
R(x) = 1800x - 23x^2 - x^3
\]
To find the maximum revenue, we first compute the derivative of $ R(x) $ with respect to $ x $, denoted $ R'(x) $.

\[
R'(x) = \frac{d}{dx}(1800x - 23x^2 - x^3)
\]
Applying the power rule for each term:
\[
R'(x) = 1800 - 46x - 3x^2
\]

### 2. Set $ R'(x) = 0 $ to find critical points
To find the critical points, set the derivative equal to zero:
\[
1800 - 46x - 3x^2 = 0
\]
Rearrange the terms:
\[
3x^2 + 46x - 1800 = 0
\]

This is a quadratic equation in $ x $. We can solve it using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where $ a = 3 $, $ b = 46 $, and $ c = -1800 $.

#### Substitute the values:
\[
x = \frac{-46 \pm \sqrt{46^2 - 4 \cdot 3 \cdot (-1800)}}{2 \cdot 3}
\]
\[
x = \frac{-46 \pm \sqrt{2116 + 21600}}{6}
\]
\[
x = \frac{-46 \pm \sqrt{23716}}{6}
\]
\[
x = \frac{-46 \pm 154}{6}
\]

This gives us two possible values for $ x $:
1. $ x = \frac{-46 + 154}{6} = \frac{108}{6} = 18 $
2. $ x = \frac{-46 - 154}{6} = \frac{-200}{6} = -33.33 $

Since $ x $ represents the number of computers, it must be non-negative. Therefore, we discard $ x = -33.33 $.

Thus, the critical point is:
\[
x = 18
\]

### 3. Verify that $ x = 18 $ maximizes the revenue
To confirm that this is a maximum, we can use the second derivative test by finding $ R''(x) $.

\[
R''(x) = \frac{d}{dx}(1800 - 46x - 3x^2)
\]
\[
R''(x) = -46 - 6x
\]

Now, substitute $ x = 18 $ into $ R''(x) $:
\[
R''(18) = -46 - 6 \cdot 18 = -46 - 108 = -154
\]
Since $ R''(18) < 0 $, the revenue function is concave down at $ x = 18 $, confirming a maximum at this point.

### 4. Find the maximum revenue
Now, substitute $ x = 18 $ back into the original revenue function $ R(x) $ to find the maximum revenue:

\[
R(18) = 1800(18) - 23(18)^2 - (18)^3
\]
Calculate each term:
\[
R(18) = 32400 - 23 \cdot 324 - 5832
\]
\[
R(18) = 32400 - 7452 - 5832
\]
\[
R(18) = 19116
\]

### Conclusion
The level of sales that maximizes revenue is:
\[
x = 18 \text{ computers}
\]
The maximum revenue is:
\[
R(x) = 19116 \text{ dollars}
\]

Would you like more details on any step? 

---

Here are 5 related questions to expand your understanding:
1. How would the results change if the revenue function had a different coefficient for the $ x^2 $ term?
2. What is the interpretation of the second derivative test in revenue maximization problems?
3. How could we verify this solution using a graphing calculator or software?
4. What other methods exist for finding the maximum of a polynomial function?
5. How would this analysis change if we were asked for the minimum revenue instead?

**Tip:** In revenue and cost functions, identifying critical points helps find the optimal points where profit or revenue peaks.

If the total revenue function for a computer is R(x) = 1800x − 23x^2 − x^3, find the level of sales, x, that maximizes revenue and find the maximum revenue in dollars.

Learn how to find the sales level that maximizes revenue for the function R(x) = 1800x − 23x^2 − x^3. This solution includes differentiation, the quadratic formula, and the second derivative test. Ideal for college calculus students studying revenue maximization.

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