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 <em>x </em>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, <em>x </em>= 30 is -($300) unexpected
- The profit when 10 more units is sold than the break-even point, x = 10 is $100
<h3>How can the given functions be used to find the profit made?</h3>
The cost is c(x) = 300 + 260•x
Revenue is r(x) = 300•x - x²
First part;
At the break even point, we have;
Which gives;
300 + 260•x = 300•x - x²
x² + 260•x - 300•x + 300 = 0
Factoring the above quadratic equation gives;
x² - 40•x + 300 = (x - 30)•(x - 10) = 0
At the break even point, <em>x </em>= 30, or <em>x</em><em> </em>= 10
The values of <em>x </em>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;
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
When the number of units sold, <em>x </em>= 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 <em>x </em>= 20, we have;
P(20) = 20•(40 - 20) - 300 = 100
- When the number of units sold, x = 20, the profit is, P(20) = $100
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