A baseball team plays in a stadium that holds 68000 spectators. With the ticket price at $11 the average attendence has been 27000. When the price dropped to $10, the average attendence rose to 34000. Assume that attendence is linearly related to ticket price.
What ticket price would maximize revenue? $
I have no idea how to go about doing this optimization question, I've been stuck on trying to find a formula for this, I've been around the internet trying to find out how to do this question. If someone can break this down for me, that would be greatly appreciated...GREATLY!
y = -7000x + 104000
the person above got you to here, which is the line function of the amount of people vs. cost of ticket
we are looking for revenue, so (number of people) x (cost of ticket)
that will get us revenue, the cost of ticket is x,
so
(-7000x + 104000)(x)
R = -7000x^2 + 104000x = Revenue
now take the derivative,
dR/dx = -14000x + 104000
then find the critical number, (what makes this equal 0)
and that should be the best answer but it is possible you would want to find out the revenue is everyone showed up and what the cost would be then, but that one is unlikely.
I tried this way several times and it is not correct. Are there any other ways to doing this problem?
Well, I can understand why you're feeling stuck. Optimization problems can be tricky sometimes. But don't worry, I'm here to help you out with a little humor!
To solve this problem, we need to find the ticket price that maximizes revenue. Revenue is calculated by multiplying the number of spectators by the ticket price.
Let's start by finding a formula for the number of spectators based on the ticket price. We know that attendance is linearly related to the ticket price, so we can use the formula:
Attendance = mx + b
where m is the rate of change in attendance with respect to ticket price and b is the intercept (attendance when ticket price is 0).
Using the given information, we can set up two equations:
27000 = m(11) + b (equation 1)
34000 = m(10) + b (equation 2)
Now, we have two linear equations and two variables (m and b). We can solve these equations simultaneously to find their values.
Let's do some algebraic calculations:
From equation 1:
m(11) + b = 27000
From equation 2:
m(10) + b = 34000
Now, let's solve these equations simultaneously. By subtracting equation 2 from equation 1, we eliminate the b term:
m(11) + b - (m(10) + b) = 27000 - 34000
11m - 10m = -7000
m = -7000
Substituting the value of m into equation 1, we get:
-7000(11) + b = 27000
-77000 + b = 27000
b = 27000 + 77000
b = 104000
Now we have the values of m and b. The equation for attendance becomes:
Attendance = -7000x + 104000
To find the ticket price that maximizes revenue, we need to multiply attendance by ticket price. We can use the revenue formula:
Revenue = Ticket Price * Attendance
Substituting the attendance equation into the revenue formula, we get:
Revenue = Ticket Price * (-7000x + 104000)
Now, we need to find the ticket price that maximizes this revenue equation. To do this, we can take the derivative of the revenue equation with respect to the ticket price and set it equal to zero:
dRevenue/dTicketPrice = 0
Solving this equation will give us the ticket price that maximizes revenue. From the revenue equation, the derivative is:
dRevenue/dTicketPrice = -7000x + 104000
Setting this equal to zero and solving for x, we find:
-7000x + 104000 = 0
-7000x = -104000
x = 104000/7000
x = 14.857
So, approximately, the ticket price that maximizes revenue is $14.86.
Now, I hope this explanation didn't drive you crazy. Remember, solving optimization problems can sometimes be as challenging as hitting a curveball! Happy calculating!
To find the ticket price that maximizes revenue, you need to understand the relationship between the ticket price and attendance, as well as how revenue is calculated.
Let's start by assuming that attendance is a linear function of ticket price. We can express attendance as a function of the ticket price using the equation of a line:
A = mP + b
Where A represents attendance, P represents the ticket price, m represents the slope of the line, and b represents the y-intercept. In this case, m is the rate at which attendance changes with respect to ticket price.
Given the information in the problem, we can determine the values of m and b.
First, we know that when the ticket price is $11, the average attendance is 27,000. Substituting these values into our equation, we get:
27,000 = 11m + b ---(1)
Similarly, when the ticket price is $10, the average attendance is 34,000:
34,000 = 10m + b ---(2)
Now we have a system of two equations (1 and 2) that we can solve simultaneously to find the values of m and b.
Subtracting equation (2) from equation (1), we get:
27,000 - 34,000 = 11m + b - 10m - b
-7,000 = m
Now, substituting this value of m into either equation (1) or (2), we can solve for b. Let's use equation (1):
27,000 = 11(-7,000) + b
27,000 = -77,000 + b
b = 27,000 + 77,000
b = 104,000
Now we can go back to our original equation A = mP + b and substitute the values we found for m and b:
A = -7,000P + 104,000
Next, we need to understand how revenue is calculated. Revenue is the product of attendance and ticket price:
R = AP
Substituting the equation for A into the equation for R:
R = (-7,000P + 104,000)P
R = -7,000P^2 + 104,000P
To find the ticket price that maximizes revenue, we need to find the maximum point of this quadratic function. The maximum point occurs at the vertex of the parabola, which can be found using the formula:
P = -b / (2a)
In this case, a = -7,000 and b = 104,000. Plugging these values into the formula, we get:
P = -(104,000) / (2*(-7,000))
P = -(104,000) / (-14,000)
P = 7.43
Therefore, the ticket price that will maximize revenue is approximately $7.43.
Please note that this calculation assumes a linear relationship between ticket price and attendance, which may not perfectly reflect the actual situation. Additionally, other factors such as the team's popularity and other external factors may influence revenue as well.
Treat it as two ordered pairs,
(11,27000) and (10,34000) and then find the equation of the line like you did in grade 9
slope = (34000-27000)/(10-11)
= -7000
y = mx + b
y = -7000x + b
using (10,34000)
34000 = 10(-7000) + b
b = 104000
y = -7000x + 104000 , y ≤68000
so at a maximum of y, y = 68000
68000 = -7000x + 104000
7000x = 36000
x = 36/7 = 5.14..
They will fill the stadium when the price of a ticket is $5.14
At that price, the revenue
= 68000(5.14) = $349,714.29