Analyze Profit Functions

Question

For a certain company, the cost function for producing x items is 

C(x)=50x+250 and the revenue function for selling x items is 

R(x)=−0.5(x−100)^2+5,000 The maximum capacity of the company is 140 items.

  

1. Assuming that the company sells all that it produces, what is the profit function? P(x)=
2. What is the domain of P(x)?
3. Hint: Does calculating P(x) make sense when x=−10 or x=1,000?
4. The company can choose to produce either 50 or 60 items. What is their profit for each case, and which level of production should they choose?
5. Profit when producing 50 items = 
6. Profit when producing 60 items =  
7. Can you explain, from our model, why the company makes less profit when producing 10 more units?

 

Answer 1

C(x) = 50x + 250

R(x) = -.5(x-100)^2 + 5000

=-.5(x^2 -200x +10000) +5000

=-x^2/2 + 100x - 5000 +5000

= -x^2/2 +100x

Profit = P(x) = R-C = -x^2/2 +50x-250

Max Profit is when P'(x) = 0

-x +50 = 0

x =50 is the profit maximizing output level

any x>50 or x<50 will have less profits

producing 60 begins to flood the market, drive down price and drive down revenue more than driving costs up.

domain is x in the interval [0,140]

profit for x=50 is -(1/2)(50^2) +50(50) -250 = 1250-250= 1,000

profit for x=60 is -(1/2)60^2 +50(60) -250 = -1800+3000-250 = 950

profit for x=-10 is (-1/2)(-10)^2 + 50(-10) -250 = -50-500-250= -800

a negative output is not realistic, but maybe it might mean the manufacturer is buying back previously sold output, costing them more profits

profit for x=1000 is (-1/2)(1000)^2 +50(1000) -50 = -500,000+50,000-50=-449,950

since the capacity is only x=140, the 1000 is not realistic, but maybe it

means 1000 over a longer period of time than the time period of max capacity for 140

profit at x=140 is (-1/2)(140)^2 +50(140) -50 = -9,800 + 7,000-50= -2,750

loss (negative profit) at x=1000 is about 164 times the loss at x=140, so

x=1000 may just be the loss in a time period of 164 times the time period of production for x=140