The weekly demand function for x units of a product sold by only one firm is p = 800 − 0.5x dollars, and the average cost of production and sale is C(with a line on top) = 300 + 2x dollars.

(a) Find the quantity that will maximize profit.

(b) Find the selling price at this optimal quantity.

(c) What is the maximum profit?

a) x = 798   is the quantity that maximizes the profit

b)po(x) = 401 $/unit

c)P(max ) = 636105 $

Step-by-step explanation:

The weekly demand function is equal to the price by unit and the Revenue function is weekly demand times quantity of units then:

R(x) = p(x) * x

R(x) = ( 800 - 0,5*x ) * x

R(x) = 800*x - 0,5*x²

The equation for the profit is:

P(x) = R(x) - C(x)

P(x) = 800*x - 0,5*x² - ( 300 + 2*x)

P(x) = 798*x - 0,5*x² - 300

a) P(x) = 798*x - 0,5*x² - 300

Taking derivatives on both sides of the equation:

P´(x) = 798 - x

If P´(x) = 0     then    798 - x = 0

x = 798   is the quantity that maximizes the profit

b) Selling price for the optimal quantity is:

p(x) = 800 - 0,5*x

p(x) = 800 - 0,5* 798

po (x) = 800 - 399

po(x) = 401 $/unit

c) Pmax = ?

P(x) = 798*x - 0,5*x² - 300    by subtitution of  x = 798

P(max) = 636804 - 399 - 300

P(max ) = 636105 $

4096 bacteria

step-by-step explanation: