Question

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?

Answer 1

Answer:

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