If C (x) is the total cost incurred in producing x units of a certain commodity, then the average cost of producing x units of the commodity is given by C¯ = C (x) x, for x > 0. The total monthly cost (in dollars) incurred by Cannon Precision Instruments for producing x units of the model M1 camera is given by C (x) = 0.0025x 2 + 80x + 10, 000. How many cameras should Cannon produce in a month in order to have the lowest possible average monthly cost?

Question

Mathematics

Alonso Nash

5 February, 15:30

If C (x) is the total cost incurred in producing x units of a certain commodity, then the average cost of producing x units of the commodity is given by C¯ = C (x) x, for x > 0. The total monthly cost (in dollars) incurred by Cannon Precision Instruments for producing x units of the model M1 camera is given by C (x) = 0.0025x 2 + 80x + 10, 000. How many cameras should Cannon produce in a month in order to have the lowest possible average monthly cost?

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Answers (1)

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Isiah Werner · Answer 1

x = 2000 cameras

Step-by-step explanation:

C (x) Total cost in producing x units

C - = C (x) / x Average cost of producing x units x > 0

Cannon Precision Instrument

C (x) Total monthly cost for producing x units of M1 cameras

is C (x) = 0.0025x² + 80x + 10000

Then average cost of producing x cameras M1 is

C - (x) = (0.0025x² + 80x + 10000) / x

C - (x) = 0.0025x + 80 + 10000/x

Taking derivatives on both sides of the equation

C-' (x) = 0.0025 - 10000/x²

Then

C-' (x) = 0

(0.0025x² - 10000) / x² = 0

0.0025x² - 10000 = 0

x² = 10000 / 0.0025 x² = 4000000

x = 2000 cameras