The cost per unit in the production of an MP3 player is $60. The manufacturer charges $90 per unit for orders of 100 or less. To encourage large orders, the manufacturer reduces the charge by $0.15 per MP3 player for each unit ordered in excess of 100 (for example, the charge is reduced to $87 per MP3 player for an order size of 120).

(a) Write the profit as a function of x.

(b) How many radios should he sell to maximize profit?

Question

Business

Janae Koch

1 February, 06:49

The cost per unit in the production of an MP3 player is $60. The manufacturer charges $90 per unit for orders of 100 or less. To encourage large orders, the manufacturer reduces the charge by $0.15 per MP3 player for each unit ordered in excess of 100 (for example, the charge is reduced to $87 per MP3 player for an order size of 120).

(a) Write the profit as a function of x.

(b) How many radios should he sell to maximize profit?

+1

Answers (1)

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Rabbit · Answer 1

Profit - 0.15x2 + 15X + 3,000

It maximize profit at 150 radios priced at 82.5

Explanation:

Total revenue will be P x Q

where P = $90-0.15X

and Q = 100 + X

TR = P x Q = (90 - 0.15X) (100 + X) =

-0.15x2 + 90X - 15X + 9,000 = - 0.15x2 + 75X + 9,000

The Cost will be C x Q = 60 x (100+X) = 60X + 6,000

Profit = TR - C = - 0.15x2 + 75X + 9,000 - 60X - 6,000 =

-0.15x2 + 15X + 3,000

As this isa quadratic function the profit it maximize at the vertex

-b/2a: - 15/2 (-0.15) = - 15/-0.3 = 50

being X the amount above 50 then

Q = 150

P = 90 - 0.15 (50) = 90 - 7.5 = 82.5