A sign-making company uses the function R(p)=−12p2+200p to model the revenue of the company, where p is the number of signs sold and R(p) is the revenue, in dollars. What is the reasonable domain and range of the polynomial function?(1 point)

Question

Responses

Domain: [0,∞);

Range: [0,∞)

Domain: left bracket 0 comma infinity right parenthesis ; Range: left bracket 0 comma infinity right parenthesis

Domain: (−∞,∞);

Range: (−∞,∞)

Domain: open paren negative infinity comma infinity close paren ; Range: open paren negative infinity comma infinity close paren

Domain: [0,20,000);

Range: [0,400)

Domain: left bracket 0 comma 20 comma 000 right parenthesis ; Range: left bracket 0 comma 400 right parenthesis

Domain: [0,400);

Range: [0,20,000]

Answer 1

To determine the reasonable domain and range of the revenue function \( R(p) = -12p^2 + 200p \), we need to analyze the function.

1. Domain:

   • The variable \( p \) represents the number of signs sold. Since the number of signs cannot be negative, the minimum value of \( p \) is 0.
   • The function is a quadratic function that opens downwards (the coefficient of \( p^2 \) is negative), meaning it reaches a maximum at some point. We should also consider where the quadratic becomes zero to understand when revenue ceases to be generated.

   To find the maximum, we can use the vertex formula \( p = -\frac{b}{2a} \) where \( a = -12 \) and \( b = 200 \): \[ p = -\frac{200}{2 \times -12} = \frac{200}{24} = \frac{25}{3} \approx 8.33 \] This implies that the maximum revenue occurs at approximately \( p \approx 8.33 \). However, revenue cannot go negative, and the function will yield zero revenue at the endpoints, which can be determined by finding the roots of the equation \( -12p^2 + 200p = 0 \): \[ p(-12p + 200) = 0 \implies p = 0 \text{ or } p = \frac{200}{12} \approx 16.67 \] Therefore, the feasible domain of \( p \) is from 0 to approximately 16.67, but since signs sold is a discrete quantity, we can represent it as \( p \in [0, 16.67] \).

2. Range:

   • To find the maximum revenue, we calculate \( R(p) \) at the vertex: \[ R\left(\frac{25}{3}\right) = -12\left(\frac{25}{3}\right)^2 + 200\left(\frac{25}{3}\right). \] Simplifying this gives the maximum revenue value. \[ R\left(\frac{25}{3}\right) = -12 \times \frac{625}{9} + \frac{5000}{3} = -\frac{7500}{9} + \frac{15000}{9} = \frac{7500}{9} \approx 833.33 \]

   Thus, revenue ranges from 0 (when no signs are sold) to approximately 833.33, which means the range is \( R \in [0, 833.33] \).

So the dimensional analysis gives us:

• Domain: \([0, 16.67]\)
• Range: \([0, 833.33]\)

However, among the options provided, the closest reasonable representation of the domain would be:

Domain: [0,∞); Range: [0,400).

Note that "Range: [0, 400)" is conservative and does not capture the actual upper bound, but it is what is provided in the options. Thus, you can summarize as follows:

Final Answer: Domain: [0,∞); Range: [0,400).