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?

Answer:

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 $

a) The quantity that will maximize profit is [tex]100 \ units[/tex].

b) The selling of at this optimal quantity is [tex]\[/tex][tex]750[/tex] per unit.

c) So, the maximum profit is [tex]\[/tex][tex]25000[/tex].

A demand function is a mathematical equation that expresses the demand for a product or service as a function of its price and other factors such as the prices of the substitutes and complementary goods, income, etc.

The given demand function,

[tex]P=800-\frac{1}{2}x[/tex]

So, the revenue function is,

[tex]R(x)=p.x\\R(x)=800-x-\frac{1}{2}x^{2}[/tex]

[tex]\Rightarrow[/tex]Average cost is

[tex]\bar{C}=300+2x[/tex]

So, the total cost function is,

[tex]C\left ( x \right )=\bar{C}.x \\ C\left ( x \right )=300x+2x^{2}[/tex]

a) Profit function:

[tex]P\left ( x \right )=R\left ( x \right )-C\left ( x \right ) \\ =800x-\frac{1}{2}x^{2}-300x-2x^{2} \\ P\left ( x \right )=500x-\frac{5}{2}x^{2}[/tex]

For maximum,

[tex]{P}'\left ( x \right )=0 \\ 500\left ( 1 \right )-\frac{5}{2}\left ( 2x \right )=0 \\ 500-5x=0 \\ x=100[/tex]

b) Selling Price:

[tex]P=800-\frac{1}{2}x \\ at \ x=100 \\ P=800-\frac{1}{2}\left ( 100 \right ) \\ =\$ 750[/tex]

c) Maximum Profit:

[tex]P\left ( x \right )=500x-\frac{5}{2}x^{2} \\ P\left ( 100 \right )=500\left ( 100 \right )-\frac{5}{2}\left ( 100 \right )^{2} \\ P\left ( 100 \right )=\$ 25000[/tex]

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