Suppose that the price per unit in dollars of a cell phone production is modeled by p = $45 − 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.
_______ thousand phones
where does 80 come from?????
no le entiendo ni madres
see last question you posted. here,
R=x(45-.0125x) then proceed as in the other problem
R=45x-.0125x^2
multiply through by 80
80R=80*45x-x^2
x^2-80*45+80R=0
for max, x=-b/2a=40(45)
véase la última pregunta informados. aquí,
R = x (45 .0125x) a continuación, proceder como en el otro problema
R = 45x-.0125x ^ 2
se multiplican a través de 80
80R = 80 * 45x-x ^ 2
x ^ 2-80 * 45 + 0 = 80R
para max, x = -b / 2a = 40 (45)
baia
To find the production level that will maximize revenue, we need to determine the value of x that will result in the maximum value of R. The revenue is given by the equation R = x * p, where x is the production level and p is the price per unit.
Substituting the expression for p into the equation for R, we have:
R = x * (45 - 0.0125x)
To find the production level that maximizes revenue, we need to find the value of x that gives the maximum value of R. One way to do this is by applying calculus. We can take the derivative of R with respect to x, set it equal to zero, and solve for x.
Let's differentiate R with respect to x:
dR/dx = 45 - 0.025x
Setting dR/dx equal to zero, we have:
45 - 0.025x = 0
Solving for x, we get:
0.025x = 45
x = 45 / 0.025
x = 1800
Therefore, the production level that will maximize revenue is 1800 thousand phones.