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. thousand phones resources ebook

Explanation:

Step1: Substitute price formula into revenue formula

Given $p = 75- 0.0125x$ and $R=x\cdot p$, then $R=x(75 - 0.0125x)=75x-0.0125x^{2}$.

Step2: Recognize the function type

$R(x)=- 0.0125x^{2}+75x$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a=-0.0125$, $b = 75$, $c = 0$.

Step3: Find the x - value of the vertex

For a quadratic function $y=ax^{2}+bx + c$, the x - coordinate of the vertex (which gives the maximum for $a<0$) is $x=-\frac{b}{2a}$. Substitute $a=-0.0125$ and $b = 75$ into the formula: $x=-\frac{75}{2\times(-0.0125)}$.

Step4: Calculate the value of x

$x=\frac{75}{0.025}=3000$.

Answer:

$3000$