The revenue from the sale of a product is given by the function Upper R equals 484 x minus x cubed. Selling how many units will give positive​ revenue?

Question

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Part 1
Selling more than
  
enter your response here and less than
  
enter your response here units will give positive revenue.

Answer 1

To find the range of units \( x \) that will yield positive revenue, we start with the revenue function:

\[ R(x) = 484x - x^3 \]

We need to determine when \( R(x) > 0 \). First, we will find the values of \( x \) where \( R(x) = 0 \):

\[ 484x - x^3 = 0 \]

We can factor out \( x \):

\[ x(484 - x^2) = 0 \]

This gives us two cases:

1. \( x = 0 \)
2. \( 484 - x^2 = 0 \)

For the second case:

\[ x^2 = 484 \]

Taking the square root:

\[ x = \sqrt{484} \quad \text{or} \quad x = -\sqrt{484} \]

Since we are dealing with selling units, we discard the negative value and take:

\[ x = 22 \]

Now we have identified the critical points where \( R(x) = 0 \):

1. \( x = 0 \)
2. \( x = 22 \)

Next, we will test intervals defined by these critical points to determine where the revenue is positive.

1. Interval \( (-\infty, 0) \): Choose \( x = -1 \): \[ R(-1) = 484(-1) - (-1)^3 = -484 + 1 = -483 \quad (\text{not positive}) \]

2. Interval \( (0, 22) \): Choose \( x = 10 \): \[ R(10) = 484(10) - (10)^3 = 4840 - 1000 = 3840 \quad (\text{positive}) \]

3. Interval \( (22, \infty) \): Choose \( x = 30 \): \[ R(30) = 484(30) - (30)^3 = 14520 - 27000 = -12480 \quad (\text{not positive}) \]

Based on these tests, we find that the revenue \( R(x) \) is positive when \( 0 < x < 22 \).

Therefore, selling more than \( 0 \) units and less than \( 22 \) units will give positive revenue:

• Selling more than: 0
• Selling less than: 22

Final answer:

• Selling more than 0 and less than 22 units will give positive revenue.