Respuesta :
Answer:
a) Price that maximize revenue P=$3
b) Price that maximize revenue (with costs) P=$3.3
Explanation:
We assume a linear demand curve as:
[tex]Q(P)=Q_0+m P[/tex]
With the data given, we can calculate the parameters m and Q0:
[tex]m=(Q_2-Q_2)/(P_2-P_1)\\\\m=(8000-10000)/(4.4-4.0)=-2000/0.4\\\\m=-5000[/tex]
[tex]Q(4)=Q_0-5,000*4=10,000\\\\Q_0=10,000+5,000*4=30,000[/tex]
Then, the demand equation becomes
[tex]Q(P)=30,000-5,000 P[/tex]
The revenue can be defined as:
[tex]R=P*Q=30,000P-5,000P^2[/tex]
To maximize the revenue, we derive and equal to zero:
[tex]dR/dP=30,000-2*5,000P=0\\\\10,000P=30,000\\\\P=3[/tex]
We now modify the revenue equation to include the fixed and variable cost, and derive again:
[tex]R=QP-1,000-0.6Q\\\\R=(30,000P-5,000P^2)-1,000-0.6(30,000-5,000P)\\\\R=-5,000P^2+(30,000-0.6(-5,000))P-(1,000+0.6*30,000)\\\\R=-5,000P^2+33,000P-19,000\\\\\\dR/dP=-10,000P+33,000=0\\\\P=33,000/10,000=3.3[/tex]
The price of a hamburger that will maximize the nightly hamburger revenue would be $ 3. In turn, if the fixed costs are $ 1000 and the variable costs per hamburger are $ 0.60, the price that would maximize profits would be $ 3.25.
Since until recently hamburgers at the city sports arena cost $ 4 each, and the food concessionaire sold an average of 10,000 hamburgers on a game night, but when the price was raised to $ 4.40, hamburger sales dropped off to an average of 8,000 per night, to find the price of a hamburger that will maximize the nightly hamburger revenue assuming a linear demand curve; and determine, if the concessionaire has a fixed cost of $ 1000 per night and the variable cost is $ 0.60 per hamburger, the price of a hamburger that will maximize the nightly hamburger profit, the following calculations must be performed:
- 0.4 = 2000
- 0.01 = X
- 0.01 x 2000 / 0.4 = X
- 50 = X
- Thus, for every penny that is increased, 50 consumers decrease.
- 8,000 x 4.4 = $ 35,200
- 10,000 x 4 = $ 40,000
- 9,950 x 4.01 = $ 39,899.50
- 10,050 x 3.99 = $ 40,099.50
- 11,250 x 3.75 = $ 42,187.50
- 12,500 x 3.50 = $ 43,750
- 13,750 x 3.25 = $ 44,687.50
- 15,000 x 3 = $ 45,000
- 15,050 x 2.99 = $ 44,999.50
- 14,950 x 3.01 = $ 44,999.50
Therefore, the price of a hamburger that will maximize the nightly hamburger revenue would be $ 3.
In turn, if the fixed costs are $ 1000 and the variable costs per hamburger are $ 0.60, the price that would maximize profits arises from subtracting the average of said cost from the previous values:
- 45,000 - (15,000 x 0.6 + 1000) = 35,000
- 44,687.5 - (13,750 x 0.6 + 1000) = 35,437.5
- 43,750 - (12,500 x 0.6 + 1000) = 35,250
Therefore, if the fixed costs are $ 1000 and the variable costs per hamburger are $ 0.60, the price that would maximize profits would be $ 3.25.
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