A company has a fixed cost of 22,000 and a production cost of $6 for each disposable camera it manufactures. A unit sels for $9. (a) What is the cost function? Cx) - 2200 + 5n (b) What is the revenue function? R(x) - 9n (c) What is the profit function? P(K) - 9n- (2200 +5n) (d) Compute the profit (loss) corresponding to the production levels of 6300, 8500, and 12,200 units, respectively. P(6300) - 3200 P(8500) - 4000 P(12,200) = 26800

Algebra for College Students
10th Edition
ISBN:9781285195780
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Jerome E. Kaufmann, Karen L. Schwitters
Chapter9: Polynomial And Rational Functions
Section9.4: Graphing Polynomial Functions
Problem 44PS: A company determines that its weekly profit from manufacturing and selling x units of a certain item...
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A company has a fixed cost of 22,000 and a production cost of $6 for each disposable camera it manufactures. A unit sels for $9.
(a) What is the cost function?
Cx) = 2200 + 5n
(b) What is the revenue function?
R(x) =
9n
(c) What is the profit function?
P(K) - 9n- (2200 + 5n)
(d) Compute the profit (loss) corresponding to the production levels of 6300, 8500, and 12,200 units, respectively.
P(6300) - 3200
P(8500) - 4000
P(12,200) = 26800
xxx
Transcribed Image Text:A company has a fixed cost of 22,000 and a production cost of $6 for each disposable camera it manufactures. A unit sels for $9. (a) What is the cost function? Cx) = 2200 + 5n (b) What is the revenue function? R(x) = 9n (c) What is the profit function? P(K) - 9n- (2200 + 5n) (d) Compute the profit (loss) corresponding to the production levels of 6300, 8500, and 12,200 units, respectively. P(6300) - 3200 P(8500) - 4000 P(12,200) = 26800 xxx
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