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?

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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.