A club currently has
300300
members who pay
$ 400$400
per month for membership dues. The club's board members want to increase monthly revenue by lowering the monthly dues in hopes of attracting new members. A market research study has shown that for each
$ 1$1
decrease in the monthly membership price, an additional
33
peoplepeople
will join the club. What price should the club charge to maximize revenue? What is the maximum revenue?
Over three hundred thousand members?
An additional thirty-three dollars?
An additional 33 people?
To maximize revenue, we need to find the price that will attract the maximum number of new members while still maintaining a reasonable monthly revenue per member.
Let's start by determining how many new members will join for a $1 decrease in the monthly membership price:
Additional members per $1 decrease = 33
To find the number of new members for different price decreases, we can divide the total decrease in membership price by $1:
Total decrease in membership price = $400 - X (where X is the price decrease)
Number of new members = (400 - X) / 1 = 400 - X
Now, we can determine the total number of members after the price decrease by adding the new members to the initial number of members:
Total number of members after price decrease = 300,300 + (400 - X) = 300,700 + 400 - X = 301,100 - X
To find the monthly revenue, we multiply the total number of members after the price decrease by the membership price:
Monthly revenue = (301,100 - X) * X
To find the maximum revenue, we need to find the value of X that maximizes the monthly revenue. We can do this by finding the derivative of the revenue function and setting it equal to zero:
d(Monthly revenue) / dX = 0
Let's differentiate the revenue function:
d(Monthly revenue) / dX = 301,100 - 2X
Setting the derivative equal to zero:
301,100 - 2X = 0
-2X = -301,100
X = 150,550
Now we have the value of X that maximizes the revenue. To find the maximum revenue, we substitute this value back into the revenue function:
Monthly revenue = (301,100 - X) * X
Maximum revenue = (301,100 - 150,550) * 150,550 = 150,550 * 150,550
Therefore, the price that the club should charge to maximize revenue is $150,550, and the maximum revenue will be $150,550 * $150,550.
To determine the price that will maximize revenue, we need to consider the relationship between price, the number of members, and revenue.
Let's start by calculating the new number of members for each decrease in price. Since the market research study indicates that for each $1 decrease in price, an additional 33 people will join the club, we can create the following relationship between price (P) and the number of members (M):
M = 300300 + 33(P - 400)
Next, we need to determine the revenue (R) generated at each price. Revenue is calculated by multiplying the price by the number of members:
R = P * M
Substituting the value of M from the previous equation into the revenue equation, we get:
R = P * (300300 + 33(P - 400))
To maximize revenue, we can simplify this equation and find its maximum by differentiating it with respect to P and setting the derivative equal to zero.
To find the maximum revenue, we take the derivative and set it equal to zero:
dR/dP = 0
300300 + 33(P - 400) + 33P = 0
300300 + 33P - 13200 + 33P = 0
66P = 13200 - 300300
66P = -287100
P = -287100 / 66 ≈ -4350
Since the price cannot be negative, we disregard this solution.
Now we can analyze the revenue generated at some critical prices:
For P = 400 (initial price):
R = 400 * 300300 ≈ $120,120,000
For P = 399:
R = 399 * (300300 + 33(399 - 400))
R ≈ $119,681,401
For P = 398:
R = 398 * (300300 + 33(398 - 400))
R ≈ $119,243,204
We can observe that the revenue decreases as the price decreases. Therefore, the maximum revenue occurs at the initial price of $400.
The maximum revenue is approximately $120,120,000.