Question

The Cost per unit of producing a radio is $60. The manufacturer charges $90 per unit for orders of 100 or less. To encourage large orders, however, the manufacturer reduces the charge by $0.10 per radio for each order in excess of 100 units. For instance, an order of 101 radios would be $89.90 per radio, and order of 102 radios would be $89.80 per radio, and so on. Find the largest order the manufacturer would allow to obtain a maximum profit.

SHOW WORK PLEASE

Answer 1

Profit function = $90 times n (for n < 100 units)
= ($90-$0.10*(n-100)) times n (for n > 100 units)

Profit maximization means, solve for n where derivative = 0.

Answer 2

would that be +100? because its only 10 cents less if you order over 100... or am i wrong? if so why??

Answer 3

Sorry, in my function above, I excluded "- 60n" for cost.

Profit(n) = n(90 - 0.1n + 10) - 60n
Profit(n) = 90n - 0.1n^2 + 10n - 60n
Derivative => -0.2n + 40 = 0
=> n = 40/0.2 = 200

Answer 4

okay could you explain how you got the equations you did? bc that seems to be my problem

Answer 5

Profit = revenue minus cost.

Revenue = price times number of units.
Cost = cost of each unit times number of units.

Price in this case is a variable depending on number of units (above 100).

Price = 90 - 0.1*(n-100). Because price decreases 10 cents per unit above 100 (according to problem description).

So, Profit = n times (90 - 0.1(n-100)) - n times $60.
= 90n - 0.1n^2 + 10n - 60n
= -0.1n^2 + 40n