For a certain company, the cost function for producing x items is C(x)=40x+150 and the revenue function for selling x items is R(x)=−0.5(x−80)2+3,200. The maximum capacity of the company is 100 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 40 or 50 items. What is their profit for each case, and which level of production should they choose?
Profit when producing 40 items =
Number
Profit when producing 50 items =
Number
Can you explain, from our model, why the company makes less profit when producing 10 more units?
To calculate the profit function, we need to subtract the cost function from the revenue function. So, the profit function is given by:
P(x) = R(x) - C(x)
Given that the revenue function is R(x) = -0.5(x-80)^2 + 3200 and the cost function is C(x) = 40x + 150, we can substitute these values into the profit function:
P(x) = (-0.5(x-80)^2 + 3200) - (40x + 150)
Simplifying further:
P(x) = -0.5(x^2 - 160x + 6400) + 3200 - (40x + 150)
P(x) = -0.5x^2 + 80x - 3200 + 3200 - 40x - 150
P(x) = -0.5x^2 + 40x - 150
The profit function is P(x) = -0.5x^2 + 40x - 150.
Now, let's determine the domain of the profit function. In this case, the maximum capacity of the company is 100 items. Therefore, the company cannot produce more than 100 items. So, the domain of P(x) should be x ≥ 0 and x ≤ 100.
For the company's profit when producing 40 items, we can substitute x = 40 into the profit function:
P(40) = -0.5(40)^2 + 40(40) - 150
P(40) = -0.5(1600) + 1600 - 150
P(40) = -800 + 1600 - 150
P(40) = 650
So, the profit when producing 40 items is 650.
Similarly, for the company's profit when producing 50 items, we can substitute x = 50 into the profit function:
P(50) = -0.5(50)^2 + 40(50) - 150
P(50) = -0.5(2500) + 2000 - 150
P(50) = -1250 + 2000 - 150
P(50) = 600
So, the profit when producing 50 items is 600.
From our model, we can see that the company makes less profit when producing 10 more units because the profit function is a quadratic equation (-0.5x^2 + 40x - 150) with a negative coefficient for the x^2 term. This means that the profit function forms a parabola opening downwards. As we increase x, the profit initially increases, reaches a maximum point, and then starts decreasing. In this case, when x increases from 40 to 50 (an increase of 10 units), the profit decreases from 650 to 600. This is because the cost of producing additional units starts outweighing the revenue generated from selling those units. Therefore, the company should choose to produce 40 items in order to maximize its profit.