To determine the range of units that give positive revenue from the revenue function \( R(x) = 225x - x^3 \), we need to find where \( R(x) > 0 \).
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Set the revenue function to zero to find critical points:
\[ 225x - x^3 = 0 \]
Factor out \( x \):
\[ x(225 - x^2) = 0 \]
This gives us two solutions:
- \( x = 0 \)
- \( 225 - x^2 = 0 \)
Now, solve for \( x \):
\[ x^2 = 225 \implies x = 15 \text{ or } x = -15 \]
Since selling a negative unit doesn't make sense, we will only consider \( x = 0 \) and \( x = 15 \).
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Determine the intervals for which \( R(x) > 0 \):
The critical points are \( x = 0 \) and \( x = 15 \). We will test the intervals determined by these points:
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Test \( x < 0 \) (e.g., \( x = -1 \)): \[ R(-1) = 225(-1) - (-1)^3 = -225 + 1 = -224 < 0 \]
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Test \( 0 < x < 15 \) (e.g., \( x = 1 \)): \[ R(1) = 225(1) - (1)^3 = 225 - 1 = 224 > 0 \]
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Test \( x > 15 \) (e.g., \( x = 16 \)): \[ R(16) = 225(16) - (16)^3 = 3600 - 4096 = -496 < 0 \]
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Conclusion:
The revenue \( R(x) \) is positive when \( 0 < x < 15 \).
Thus, selling more than 0 and less than 15 units will give positive revenue.
Final Answer:
- Selling more than \( 0 \) and less than \( 15 \) units will give positive revenue.