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 ≤ y ≤ 125000 an appropriate range for function r? explain your reasoning.

Explanation:

Step1: Analyze the revenue function

The revenue function is \( r(x)=x(14000 - 2x) \), which simplifies to \( r(x)=-2x^{2}+14000x \). This is a quadratic function in the form \( ax^{2}+bx + c \) with \( a=-2 \), \( b = 14000 \), and \( c = 0 \). Since \( a<0 \), the parabola opens downwards, and the vertex represents the maximum point.

Step2: Find the vertex of the parabola

The x - coordinate of the vertex of a quadratic function \( y = ax^{2}+bx + c \) is given by \( x=-\frac{b}{2a} \). Substituting \( a=-2 \) and \( b = 14000 \) into the formula, we get \( x=-\frac{14000}{2\times(-2)}=\frac{14000}{4} = 3500 \).

Step3: Analyze the domain of x (price per phone)

We know that the number of phones sold \( 14000 - 2x\geq0 \) (because we can't sell a negative number of phones) and \( x\geq0 \) (price can't be negative). Solving \( 14000 - 2x\geq0 \) gives \( 2x\leq14000 \) or \( x\leq7000 \). Also, from the vertex, we know that the maximum revenue occurs at \( x = 3500 \).

Step4: Analyze the range of r(x)

First, find the maximum revenue by substituting \( x = 3500 \) into the revenue function: \( r(3500)=-2\times(3500)^{2}+14000\times3500=-2\times12250000 + 49000000=-24500000+49000000 = 24500000 \). When \( x = 0 \), \( r(0)=0 \), and when \( x = 7000 \), \( r(7000)=7000\times(14000 - 2\times7000)=7000\times(14000 - 14000)=0 \). Since the parabola opens downwards, the range of the function \( r(x) \) is from \( 0 \) to \( 24500000 \). The given range \( 0\leq y\leq25000000 \) is appropriate because the actual maximum revenue is \( 24500000 \), which is less than \( 25000000 \), and the minimum revenue (when no phones are sold or when the price is so high that no phones are sold) is \( 0 \). Also, the domain of \( x \) (price) is between \( 0 \) and \( 7000 \), and the range of the revenue function is between \( 0 \) and \( 24500000 \), so \( 0\leq y\leq25000000 \) is a reasonable approximation as it covers the entire range of possible revenues (from \( 0 \) up to just above the maximum revenue of \( 24500000 \)).

Answer:

Yes, \( 0\leq y\leq25000000 \) is an appropriate range for the function \( r(x) \). The revenue function is a downward - opening parabola with a maximum revenue of \( 24500000 \) (at \( x = 3500 \)) and a minimum revenue of \( 0 \) (when \( x = 0 \) or \( x = 7000 \)). Since \( 25000000 \) is slightly greater than the actual maximum revenue of \( 24500000 \), it provides a reasonable upper bound for the range of the revenue function.