Question

a company finds that if it charges x dollars for a cell phone, it can expect to sell 1000 + 2x phones. the company uses the function r defined by r(x) = x(1,000 - 2x) to model the expected revenue, in dollars, from selling the cell phones at x dollars each.
is 0 ≤ x ≤ 600 an appropriate domain for function r, assuming the company wants to make money? explain your reasoning.

Explanation:

Step1: Analyze Revenue Function

The revenue function is \( r(x) = x(1000 - 2x) \). For the company to make money, revenue \( r(x)>0 \) and the number of phones sold \( 1000 - 2x\geq0 \) (can't sell negative phones), and \( x\geq0 \) (price can't be negative).

Step2: Solve \( r(x)>0 \)

\( x(1000 - 2x)>0 \). The roots are \( x = 0 \) and \( 1000 - 2x = 0\Rightarrow x = 500 \). The quadratic opens downward (coefficient of \( x^2 \) is \( -2 \)), so \( r(x)>0 \) when \( 0 < x < 500 \). Also, \( x\geq0 \) and \( 1000 - 2x\geq0\Rightarrow x\leq500 \). But the question asks about \( 0\leq x\leq600 \). At \( x = 600 \), \( 1000 - 2(600)= - 200 \) (negative phones sold, invalid). At \( x = 500 \), \( 1000 - 2(500)=0 \) (revenue \( 0 \)). So valid domain for making money is \( 0\leq x\leq500 \). But the question is if \( 0\leq x\leq600 \) is appropriate. Since at \( x > 500 \), phones sold are negative (not possible), so \( 0\leq x\leq600 \) includes \( x > 500 \) where model is invalid. But wait, maybe the problem's "make money" is revenue > 0. So \( 0 < x < 500 \). But the domain \( 0\leq x\leq600 \): check \( x = 0 \) (revenue 0, not making money), \( x = 600 \) (revenue negative phones, invalid). But maybe the company considers price from 0 to 600, even if at 600 no phones sold. But for "making money", revenue > 0, so \( 0 < x < 500 \). However, the domain \( 0\leq x\leq600 \): is it appropriate? Let's see the function \( r(x)= - 2x^2 + 1000x \), a quadratic. The domain for the model (regardless of making money) would have \( 1000 - 2x\geq0\Rightarrow x\leq500 \), so \( 0\leq x\leq500 \) is appropriate for the model (since number of phones can't be negative). But the question says "assuming the company wants to make money". So \( 0\leq x\leq600 \): at \( x > 500 \), revenue is negative (since \( r(x) \) is negative there, as quadratic opens down), so they can't make money there. So \( 0\leq x\leq600 \) is not appropriate because for \( x > 500 \), the number of phones sold is negative (invalid) and revenue is negative (not making money). Wait, maybe I misread. The problem says "is \( 0\leq x\leq600 \) an appropriate domain for function \( r \), assuming the company wants to make money". To make money, \( r(x)>0 \). \( r(x)>0 \) when \( 0 < x < 500 \). So \( 0\leq x\leq600 \) includes \( x\geq500 \) where \( r(x)\leq0 \) (not making money), so it's not appropriate. But let's check the function's domain without making money: the model \( r(x)=x(1000 - 2x) \) has \( 1000 - 2x\geq0\Rightarrow x\leq500 \) (since number of phones can't be negative), so domain should be \( 0\leq x\leq500 \). So \( 0\leq x\leq600 \) is not appropriate because at \( x > 500 \), the number of phones sold is negative, which doesn't make sense in the context of the model (you can't sell a negative number of phones), so the domain should be restricted to where the number of phones sold is non - negative, i.e., \( 1000 - 2x\geq0\Rightarrow x\leq500 \), so \( 0\leq x\leq500 \) is appropriate, not \( 0\leq x\leq600 \).

Answer:

To determine if \( 0\leq x\leq600 \) is appropriate for \( r(x)=x(1000 - 2x) \) (revenue, \( x \) = price) when the company wants to make money:

1. Model Constraints: The number of phones sold is \( 1000 - 2x \), which must be non - negative (\( 1000 - 2x\geq0\Rightarrow x\leq500 \)) (can’t sell negative phones). Also, price \( x\geq0 \).
2. Revenue for Making Money: To make money, \( r(x)>0 \). Solving \( x(1000 - 2x)>0 \), the roots are \( x = 0 \) and \( x = 500 \). The quadratic \( r(x)=-2x^{2}+1000x \) opens downward, so \( r(x)>0 \) when \( 0 < x < 500 \).
3. Analyze \( 0\leq x\leq600 \): For \( x > 500 \) (e.g., \( x = 600 \)), \( 1000 - 2x=1000 - 1200=-200 \) (negative phones sold, invalid) and \( r(x)=600\times(-200)<0 \) (no revenue, or negative revenue). Thus, \( 0\leq x\leq600 \) includes values where the model (number of phones) is invalid and revenue is non - positive (not making money). So \( 0\leq x\leq600 \) is not an appropriate domain for the function \( r \) when the company wants to make money, as it includes \( x > 500 \) where the number of phones sold is negative (illogical) and revenue is non - positive.

(If the question was about the model’s domain without “making money” context, the model’s domain (for valid number of phones) is \( 0\leq x\leq500 \), so \( 0\leq x\leq600 \) is still inappropriate as it includes \( x > 500 \) with negative phone sales.)