Respuesta :
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].
Demand Function:
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]
Learn more about of Demand Function: https://brainly.com/question/25938253
