For a certain company, the cost function for producing x items is C(x)=30x+150 and the revenue function for selling x items is R(x)=−0.5(x−90)2+4,050. The maximum capacity of the company is 130 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 60 or 70 items. What is their profit for each case, and which level of production should they choose?
Profit when producing 60 items =
Number


Profit when producing 70 items =
Number


Can you explain, from our model, why the company makes less profit when producing 10 more units?

Respuesta :

Profit, revenue and cost are related, and can be calculated from one another.

  • The profit function is [tex]\mathbf{P(x) =-0.5(x -90)^2 - 30x + 3900}[/tex]
  • The domain of the profit function is x > 0
  • The profit when producing 60 items is 1650
  • The profit when producing 70 items is 1600

The cost function is given as:

[tex]\mathbf{C(x) =30x + 150}[/tex]

The revenue function is given as:

[tex]\mathbf{R(x) =-0.5(x -90)^2 + 4050}[/tex]

(a) Calculate the profit function

This is calculated using:

[tex]\mathbf{P(x) = R(x) - C(x)}[/tex]

So, we have:

[tex]\mathbf{P(x) =-0.5(x -90)^2 + 4050 - 30x - 150}[/tex]

Evaluate like terms

[tex]\mathbf{P(x) =-0.5(x -90)^2 - 30x + 3900}[/tex]

(b) The domain of the profit function

When profit is 0 or less, then it becomes no profit.

Hence, the domain of the profit function is x > 0

(c) The profit for 60 and 70 items

Substitute 60 and 70 for x in P(x)

[tex]\mathbf{P(60) =-0.5(60 -90)^2 - 30 \times 60 + 3900}[/tex]

[tex]\mathbf{P(60) =1650}[/tex]

The profit when producing 60 items is 1650

[tex]\mathbf{P(70) =-0.5(70 -90)^2 - 30 \times 70 + 3900}[/tex]

[tex]\mathbf{P(70) = 1500}[/tex]

The profit when producing 70 items is 1600

(d) Why producing 10 more units less profit

When a function reaches the optimal value, the value of the function begins to reduce.

This means that, producing 10 more units takes the profit function beyond its maximum point.

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