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Owners of a car rental company that charges customers between $65 and $200 per day have determined that the number of cars rented per day n can be modeled by the linear function n(p)=800−4p, where p is the daily rental charge. How much should the company charge each customer per day to maximize revenue? Do not include units or a dollar sign in your answer.

Respuesta :

Answer:

maximize revenue is ( 100 , 40000 )

Explanation:

given data

charges customers between = $65 and $200 per day

linear function n(p) = 800−4p

p = daily rental charge

to find out

company charge each customer per day to maximize revenue

solution

we know that Revenue = n(p)*p

so given equation

R(p) = 800 - 4p × p

R(p) = 800p - 4p²         ...................1

so now find critical point here that is

f' (p) = [tex]\frac{d}{dp}[/tex] 800p - 4p²

= [tex]\frac{d}{dp}[/tex] 800p - [tex]\frac{d}{dp}[/tex] 4p²

= 800 (1)  - 4 [tex]2p^{2-1}[/tex]

800 - 8p

solve it 800 - 8p  = 0

p = 100

now put extreme point x =  100 into equation 1

y = 800 p - 4p²

y = 800 (100) - 4 (100)²

y = 40000

so maximize revenue is ( 100 , 40000 )

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