Match club charges annual membership dues of 100 pesos per member, less 0.50 cantavo for each member over 600 and plus 0.50 centavo for each member less than 600. How many members should give the club most revenue from annual dues?
they get a credit of .5centavo for each member over 600, and a .5centavo for each memember less than 600. If that is true, they are getting zero net adjustments for all members from 0 to infinity.
revenue=100*members
max revenue is when members = infinity.
for x members (assuming you meant 50 centavos, not 0.50 centavos)
r(x) =
x(100+.5(600-x)) for x < 600
x(100-.5(x-600)) for x > 600
These are really the same function:
r(x) = 400x - 0.5x^2
maximum revenue is at x = 400
To determine the number of members that will generate the most revenue from annual dues, we can analyze the effect of the membership dues on the revenue at various membership levels.
Let's denote the number of members as "x".
For each member, the annual membership dues are 100 pesos minus 0.50 centavo for each member over 600, plus 0.50 centavo for each member less than 600.
Therefore, the annual membership dues for x members can be represented as:
Dues = 100x - 0.50(x-600) + 0.50(600 - x)
Expanding and simplifying the equation:
Dues = 100x - 0.50x + 300 + 0.50x - 0.50x
Dues = 100x + 300
Since we are trying to maximize revenue, we can consider the equation as the revenue function.
To find the maximum revenue, we can use calculus by taking the derivative of the revenue function with respect to x and setting it equal to zero:
d(revenue)/dx = 100
Setting this equal to zero, we get:
100 = 0
This indicates that the revenue is constant, and the maximum revenue will be obtained at any value of x.
Therefore, the number of members that should give the club the most revenue from annual dues is dependent on other factors not provided in the given information.
To determine the number of members that would generate the most revenue from annual dues for the club, we can analyze the given information.
Let's assume that the number of members in the club is "x."
According to the given information, the club charges 100 pesos per member. Additionally, there is a discount of 0.50 centavo for each member over 600, and an increase of 0.50 centavo for each member less than 600.
To determine which value of "x" would maximize revenue, we need to find the point where the total revenue based on the given constraints is the highest.
Revenue = (100 - 0.50 * (x - 600)) * x if x > 600
Revenue = (100 + 0.50 * (600 - x)) * x if x < 600
Now, to find the maximum revenue, we will compare the two equations above.
For x > 600:
Revenue = (100 - 0.50 * (x - 600)) * x
= (100 - 0.50x + 300) * x
= (400 - 0.50x) * x
= 400x - 0.50x^2
For x < 600:
Revenue = (100 + 0.50 * (600 - x)) * x
= (100 + 0.50 * (600 - x)) * x
= (100 + 300 - 0.50x) * x
= (400 - 0.50x) * x
= 400x - 0.50x^2
The revenue functions for both scenarios are the same: 400x - 0.50x^2.
Now, to find the number of members that will give the club the most revenue, we need to find the maximum point of this quadratic equation.
The maximum point of a quadratic equation in the form of ax^2 + bx + c can be found using the formula: x = -b/(2a).
In this case, a = -0.50 and b = 400.
x = -(400)/(2 * -0.50)
x = -(400)/-1
x = 400
So, the number of members that will generate the most revenue for the club from annual dues is 400 members.