Question 1193131: 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−130)2+8,450. The maximum capacity of the company is 170 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)= Preview Change entry mode .
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 math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Question 1) What is the profit function?
R(x) = Revenue = -0.5(x-130)^2 + 8450
C(x) = Cost = 50x+250
Profit = Revenue - Cost
Profit = (Amount taken in) - (Amount spent)
P(x) = R(x) - C(x)
P(x) = [ R(x) ] - [ C(x) ]
P(x) = [ -0.5(x-130)^2 + 8450 ] - [ 50x+250 ]
P(x) = -0.5(x^2-260x+16900) + 8450 - (50x+250)
P(x) = -0.5x^2+130x-8450 + 8450 - 50x-250
P(x) = -0.5x^2+(130x-50x)+(-8450 + 8450-250)
P(x) = -0.5x^2 + 80x - 250
This plots a parabola which opens downward; hence, it produces a highest point (aka vertex) to quickly visually convey the max profit.
At the end of the day, this is what the boss cares about (assuming they don't want to bother with the intricate details of the math).
Side note: The breakeven points occur at the x intercepts when P(x) = 0. This is when the revenue and cost are equal.
Another note: The term "sales" is often used interchangeably with "revenue". Example: "The company's sales last year were $100 million".
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Question 2) What is the domain of P(x)?
x = number of items sold
It doesn't make sense to sell a negative amount of items, nor to sell a fractional amount of items.
The smallest x can be is x = 0
The largest is x = 170 because this is the manufacturing plant's max capacity
Therefore, the domain is the set of integers x such that 
x is an integer between 0 and 170, inclusive of both endpoints.
We can express this in interval notation to say [0, 170].
The square brackets mean to include each endpoint.
You could also use roster notation to write {0, 1, 2, 3, ..., 168, 169, 170}
The triple dots tell the reader to keep the pattern going of adding 1 each time until reaching 170. This is of course much preferable instead of writing literally every integer from 0 to 170.
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Question 3) 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?
Plug x = 80 into the P(x) function and compute.
P(x) = -0.5x^2 + 80x - 250
P(80) = -0.5(80)^2 + 80(80) - 250
P(80) = 2950
Selling 80 items yields a profit of $2950
Repeat for x = 90
P(x) = -0.5x^2 + 80x - 250
P(90) = -0.5(90)^2 + 80(90) - 250
P(90) = 2900
Selling 90 items yields a profit of $2900, which is less than the $2950 found earlier.
The company should produce and sell 80 items.
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Question 4) Can you explain, from our model, why the company makes less profit when producing 10 more units?
The equation y = -0.5x^2 + 80x - 250 is in the format y = ax^2+bx+c
The vertex is (h,k)
h = -b/(2a)
h = -80/(2(-0.5))
h = 80
The vertex occurs at h = 80, meaning the highest value P(x) happens when x = 80 items are sold.
Anything beyond this point will have the profit trend downhill.
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