Asked by Bri
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
_______ thousand phones
Answers
Answered by
bobpursley
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)
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)
Answered by
pila
no le entiendo ni madres
Answered by
pila
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
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
Answered by
Anonymous
3600
Answered by
Ethan
where does 80 come from?????
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