The number of cameras produces since May at a plant can be modeled by the function N (m) = 50m+600 and the cost per camera can be modeled by C (m) = 4m^2-15m+45, where m is the number of months since May. According to this model, what is the total amount of revenue generated by the plant's camera production on September?

Question

Chubs

Mathematics

28 February, 18:16

The number of cameras produces since May at a plant can be modeled by the function N (m) = 50m+600 and the cost per camera can be modeled by C (m) = 4m^2-15m+45, where m is the number of months since May. According to this model, what is the total amount of revenue generated by the plant's camera production on September?

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Answers (2)

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Tanner Nixon · Answer 1

39,200

Step-by-step explanation:

September is 4 months after may, so m = 4

No. of cameras:

N (4) = 50 (4) + 600 = 800

Cost per camera:

C (4) = 4 (4²) - 15 (4) + 45 = 49

Total cost:

N * C

800 * 49

39200

Adriana Baldwin · Answer 2

39200

Step-by-step explanation:

Revenue = number of cameras * cost

R (m) = N (m) * C (m)

= (50m+600) * (4m^2-15m+45)

The months since may is June July August September = 4 months

R (4) = (50*4+600) * (4*4^2-15*4+45)

(200+600) * (64-60+45)

800*49

39200