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You are currently evaluating your business and trying to decide how much you need to sell to make a profit. Choose one of the following options for your cost and revenue functions. The variable, x, represents the number of units sold.
c(x)=300+260x
r(x)=300x-xsquared

For the option you chose, find the value(s) of x (the number of units sold) to break-even. Show all your work by typing it in or uploading a picture of your handwritten work. What is your profit function, P(x)? What is your profit when you sell 10 more than a break-even point? Is that what you expected? Show all your work

Sagot :

From the given functions, of the cost, c(x) = 300 + 260•x, and revenue, r(x) = 300•x - x², we have;

First part;

The values of x to break-even are;

x = 30, or x = 10

Second part;

The profit function, P(x) is presented as follows;

P(x) = x•(40 - x) - 300

Third part;

  • The profit (loss) when 10 more units is sold than the break-even point, x = 30 is -($300) unexpected

  • The profit when 10 more units is sold than the break-even point, x = 10 is $100

How can the given functions be used to find the profit made?

The cost is c(x) = 300 + 260•x

Revenue is r(x) = 300•x - x²

First part;

At the break even point, we have;

  • c(x) = r(x)

Which gives;

300 + 260•x = 300•x - x²

x² + 260•x - 300•x + 300 = 0

  • x² - 40•x + 300 = 0

Factoring the above quadratic equation gives;

x² - 40•x + 300 = (x - 30)•(x - 10) = 0

At the break even point, x = 30, or x = 10

The values of x at the break even point are;

  • x = 30 units sold
  • x = 10 units sold

Second part;

Profit = Revenue - Cost

The profit function, P(x), is therefore;

P(x) = r(x) - c(x)

Which gives;

P(x) = (300•x - x²) - (300 + 260•x)

P(x) = 300•x - x² - 300 - 260•x

P(x) = 300•x - 260•x - x² - 300

P(x) = 40•x - x² - 300

The profit function is therefore;

  • P(x) = x•(40 - x) - 300

Third part;

When 10 more units are sold than the break even point, we have;

x = 30 + 10 = 40 or x = 10 + 10 = 20

The profit at x = 40 or x = 20 are;

P(40) = 40•(40 - 40) - 300 = -300

  • P(40) = -($300)

When the number of units sold, x = 40, the profit is, P(40) = -($300) unexpected loss

  • The profit (loss) when the number of units sold increases to 40, of -($300) is unexpected.

At x = 20, we have;

P(20) = 20•(40 - 20) - 300 = 100

  • P(20) = $100

  • When the number of units sold, x = 20, the profit is, P(20) = $100

Learn more about functions here:

https://brainly.com/question/88581

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