Question 1167022: im sory didnt realize i didnt include the actual function here it is.
For a certain company, the cost function for producing x items is C(x)=30x+250 and the revenue function for selling x items is R(x)=−0.5(x−110)2+6,050. The maximum capacity of the company is 140 items.
The profit function P(x) is the revenue function R(x) (how much it takes in) minus the cost function C(x) (how much it spends). In economic models, one typically assumes that a company wants to maximize its profit, or at least make a profit!
Answers to some of the questions are given below so that you can check your work.
Assuming that the company sells all that it produces, what is the profit function?
P(x)=
Hint: Profit = Revenue - Cost as we examined in Discussion 3.
What is the domain of P(x)?
Hint: Does calculating P(x) make sense when x=−10 or x=1,000?
The company can choose to produce either 80 or 90 items. What is their profit for each case, and which level of production should they choose?
Profit when producing 80 items =
Profit when producing 90 items =
Can you explain, from our model, why the company makes less profit when producing 10 more units?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! −0.5(x−110)^2+6,050-(30x+250) is the profit function
This is -0.5(x^2-220x+12100)+6050-30x-250
or -0.5x^2+110x-6050+6050-30x-250
or -0.5x^2+80x-250
The domain is all x but -units make no sense and 1000 units is a large loss.
At 80 units the profit is -3200+6400-250 or $2950
at 90 units it is $2800
The maximum is when x=-b/2a or -80/-1 or 80 units.
With squared functions as production (in this instance) increases, there could be decreasing revenue simply because the company may oversupply and have to either store or sell at a lower price. This is an economics question rather than a math question.
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