.
Suppose that the price per unit in dollars of a cell phone production is modeled by p = $35 − 0.0125x,
where x is in thousands of phones produced, and the revenue represented by thousands of dollars is R = x · p.
Find the production level that will maximize revenue.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`
According to the condition, the revenue (measured in thousands of dollars) is
R(x) = x*(35-0.0125x),
where x is the number of the phones measured in thousand of units.
So, you need to find the maximum of this quadratic function
R(x) = -0.0125x^2 + 35x.
The maximum is achieved at x = ( referring to the general form of a quadratic function q(x) = ),
which at given conditions is x = = = 1400.
So, the maximum is achieved at the production level 1400 thousand of phone units .
The maximum revenue is the value R(x) at this value of x:
= R(1400) = = 24500 thousands of dollars.
Answer. The maximum revenue is 24500 thousands of dollars achieved at the production level of 1400 thousand of phone units .
Solved.
To see other similar solved problems, see the lesson
- Using quadratic functions to solve problems on maximizing revenue/profit
in this site.
On finding the maximum/minimum of a quadratic function see the lessons
- HOW TO complete the square to find the minimum/maximum of a quadratic function
- Briefly on finding the minimum/maximum of a quadratic function
- HOW TO complete the square to find the vertex of a parabola
- Briefly on finding the vertex of a parabola
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this textbook under the topic "Finding minimum/maximum of quadratic functions".