Answer:
Step-by-step explanation:
From the information given:
p = 70 - 0.02q
when:
q = 3000
p = 70 - 0.02 (3000)
p = 70 - 60
p = 10 dollars
The total revenue = price × number of people
= 10 × 3000
= 30000 dollars
If the price is $20
20 = 70 - 0.02q
0.02q = 70 - 20
0.02q = 50
q = 50/0.02
q = 2500
∴
If price is $20, The total revenue = 20 × 2500
= 50000 dollars
b)
The revenue function R = price × attendance
R = p × q
R = (70 - 0.02q)q
R = 70q - 0.02q²
c)
R(q) = 70q - 0.02q²
Taking the differential of the value on the right hand side.
R'(q) = 70 × (1) - 0.02 × (2q)
R'(q) = 70 - 0.04q
If we set R'(q) = 0
70 - 0.04q = 0
70 = 0.04q
q = 70/0.04
q = 1750 (critical point)
R"(q) = 0 - 0.04 × 1
R"(q) = -0.04
R"(1750) = -0.04 < 0
∴
R is maximum when q = 1750
d)
At q = 1750
p = 70 - 0.02(q)
p = 70 - 0.02(1750)
p = 70 - 35
p = 35 dollars
e)
R(q) = 70q - 0.02q² because R is max at q = 1750
R(1750) = 70 × 1750 - 0.02 × (1750)²
R_(max) = 122500 - 61250
R_(max) = 61250 dollars
Given that the actual price is 20 dollars when the price is changed to 35 dollars, we have a maximum revenue of:
Suppose:
p = 20 and q = 2500 (from a)
Then;
R = p × q
R = 20 × 2500
R = 50000 dollars
If R = 35, then q = 1750
R_{max} = 35 × 1750
R_{max} = 61250 dollars
∴
Profit = R_{max} - R
Profit = 61250 - 50000
Profit = 11250 dollars
