Question
A hardware store uses the function P(x)=−x2+400x−1,500 to model the profit of the store, where x is the number of units sold and P(x) is the profit in dollars. What is the reasonable domain and range of the polynomial function?(1 point) Responses Domain: {x:x∈Z,x≥0}; Range: {y:y∈R,y≤38,500} Domain: start set x such that x is in cap z comma x is greater than or equal to 0 end set ; Range: start set y such that y is in cap r comma y is less than or equal to 38 comma 500 end set Domain: {x:x∈Z,x≥0}; Range: {y:y∈R} Domain: start set x such that x is in cap z comma x is greater than or equal to 0 end set ; Range: start set y such that y is in cap r end set Domain: {x:x∈R;x≥0} ; Range: {y:y∈R,y≤38,500} Domain: {x:x∈R;x≥0} ; Range: start set y such that y is in cap r comma y is less than or equal to 38 comma 500 end set Domain: {x:x∈R}; Range: {y:y∈R}
Answers
GPT-4o mini
To determine the reasonable domain and range of the profit function \( P(x) = -x^2 + 400x - 1500 \), we must consider the context of the problem.
1. **Domain**:
- The variable \( x \) represents the number of units sold, which cannot be negative. Therefore, the reasonable domain is \( x \geq 0 \).
- Since we are generally dealing with whole units sold, we also consider \( x \) to be an integer (whole number). Thus, the domain can be stated as \( \{x : x \in \mathbb{Z}, x \geq 0\} \).
2. **Range**:
- To find the range of the profit function, we first note that the function is a quadratic equation that opens downward (as the coefficient of \( x^2 \) is negative).
- The vertex of this parabola will give us the maximum profit. The \( x \)-coordinate of the vertex can be calculated using the formula \( x = -\frac{b}{2a} \), where \( a = -1 \) and \( b = 400 \):
\[
x = -\frac{400}{2 \cdot -1} = 200
\]
- We can find the maximum profit by substituting \( x = 200 \) back into the profit function:
\[
P(200) = -200^2 + 400 \cdot 200 - 1500 = -40000 + 80000 - 1500 = 38500
\]
- The maximum profit is $38,500, which occurs when 200 units are sold.
From this analysis, the reasonable range for \( P(x) \) (profit) is \( P(x) \leq 38,500 \). Since \( P(x) \) can also be negative if too many units are sold, we use \( y \in \mathbb{R} \) but specify that \( y \leq 38,500 \).
Combining these findings:
- **Domain**: \( \{x : x \in \mathbb{Z}, x \geq 0\} \)
- **Range**: \( \{y : y \in \mathbb{R}, y \leq 38,500\} \)
So the correct choice is:
**Domain: {x:x∈Z,x≥0}; Range: {y:y∈R,y≤38,500}**
1. **Domain**:
- The variable \( x \) represents the number of units sold, which cannot be negative. Therefore, the reasonable domain is \( x \geq 0 \).
- Since we are generally dealing with whole units sold, we also consider \( x \) to be an integer (whole number). Thus, the domain can be stated as \( \{x : x \in \mathbb{Z}, x \geq 0\} \).
2. **Range**:
- To find the range of the profit function, we first note that the function is a quadratic equation that opens downward (as the coefficient of \( x^2 \) is negative).
- The vertex of this parabola will give us the maximum profit. The \( x \)-coordinate of the vertex can be calculated using the formula \( x = -\frac{b}{2a} \), where \( a = -1 \) and \( b = 400 \):
\[
x = -\frac{400}{2 \cdot -1} = 200
\]
- We can find the maximum profit by substituting \( x = 200 \) back into the profit function:
\[
P(200) = -200^2 + 400 \cdot 200 - 1500 = -40000 + 80000 - 1500 = 38500
\]
- The maximum profit is $38,500, which occurs when 200 units are sold.
From this analysis, the reasonable range for \( P(x) \) (profit) is \( P(x) \leq 38,500 \). Since \( P(x) \) can also be negative if too many units are sold, we use \( y \in \mathbb{R} \) but specify that \( y \leq 38,500 \).
Combining these findings:
- **Domain**: \( \{x : x \in \mathbb{Z}, x \geq 0\} \)
- **Range**: \( \{y : y \in \mathbb{R}, y \leq 38,500\} \)
So the correct choice is:
**Domain: {x:x∈Z,x≥0}; Range: {y:y∈R,y≤38,500}**