Question

suppose that the price per unit in dollars of a cell phone production is modeled by ( p = 75 - 0.0125x ), where ( x ) is in thousands of phones produced, and the revenue represented by thousands of dollars is ( r = xcdot p ). find the production level that will maximize revenue.

given a formula for price per unit of a cell phone production ( p ) in terms of the number of cell phones produced ( x ) and a general revenue formula ( r = x p ), what equation models the revenue of cell phone production? notice that the cell phone production revenue equation is a downward opening parabola. which coordinate of the vertex gives the number of cell phones produced when the revenue reaches the maximum value?

Explanation:

Step1: Substitute \( p \) into revenue formula

Given \( p = 75-0.0125x \) and \( R=x\cdot p \), substitute \( p \) into \( R \):
\( R=x(75 - 0.0125x)=75x-0.0125x^{2} \).

Step2: Use vertex formula for quadratic function

For a quadratic function \( y = ax^{2}+bx + c \) (here \( a=-0.0125\), \( b = 75\), \( c = 0 \)), the \( x \)-coordinate of the vertex is \( x=-\frac{b}{2a} \).
Substitute \( a=-0.0125 \) and \( b = 75 \) into \( x=-\frac{b}{2a} \):

$$
LATEXBLOCK0
$$

Answer:

\(3000\)