I am having a hard time figuring this out. I know I need to find the vertex from the vertex form equation but I am having a hard time figuring out the equation.
A caterer will charge a certain club $20 per member to cater a dinner if 150 members attend. However, the cost per member will be reduced by 80 cents for each member attending in excess of 150. What is the maximum revenue that the caterer can make? If you were the caterer, what upper limit would you set for the number of members that can attend the dinner?
So far I have this equation:
price per member = 20 - 0.80 (y - 150)
Ok. So, the revenue is the price times the attendance:
(20-.80(y-150))(y) for y >= 150
You sure there's no typo? I get a maximum revenue for y < 150.
Maybe the first 150 still get charged $20, but the extras get charged less. In that case, for y>150, the revenue is
20*150 + (20-.8(y-150))(y-150)
Then the maximum is at y=162 and it drops back to 3000 at y=175.
To find the vertex form equation, we can start with the given equation:
Price per member = 20 - 0.80(y - 150)
In this equation, "y" represents the number of members attending the dinner. To find the maximum revenue, we need to consider the relationship between the number of members and the cost per member.
The maximum revenue occurs when the number of members attending maximizes the cost per member. To determine this, we need to find the vertex of the quadratic equation.
To convert the equation into vertex form, we'll first expand and simplify:
Price per member = 20 - 0.80y + 0.80(150)
Price per member = 20 - 0.80y + 120
Price per member = -0.80y + 140
Now, we can rewrite the equation in vertex form by completing the square:
Price per member = -0.80(y - 175) + 140
The vertex form equation is: Price per member = -0.80(y - h) + k, where the vertex is at (h, k).
From this equation, we can see that the vertex occurs when (y - h) is equal to 0. Therefore, the maximum revenue occurs when y = 175 (150 + h), and the cost per member is given by k = 140.
So the maximum revenue that the caterer can make is achieved when 175 members attend, with a cost per member of $140.
As for setting an upper limit for the number of members that can attend the dinner, it depends on the caterer's goals and capacity. Since the maximum revenue occurs at 175 members, setting a limit slightly below that number, such as 170 or 172 members, could be a reasonable choice. Ultimately, the caterer should consider factors such as space, resources, and profitability to determine the ideal upper limit for attendance.
To find the maximum revenue that the caterer can make, we need to determine the number of members that maximizes the revenue.
In this case, the revenue is calculated by multiplying the price per member by the number of members attending the dinner. The equation you have correctly represents the price per member:
price per member = 20 - 0.80(y - 150)
To determine the revenue equation, we multiply the price per member by the number of members attending, y:
revenue = (20 - 0.80(y - 150)) * y
To find the maximum revenue, we need to find the vertex of the revenue equation. The vertex form for a quadratic equation is given as:
y = a(x - h)^2 + k
Where (h, k) is the vertex of the parabola. In our case, we can rewrite the revenue equation in vertex form by completing the square:
revenue = -0.80(y - 150)^2 + 20(y - 150)
Now we can see that the coefficient of (y - 150)^2 is -0.80 and the coefficient of (y - 150) is 20. The vertex of the parabola occurs at y = 150, so let's substitute that value into the equation:
revenue = -0.80(150 - 150)^2 + 20(150 - 150)
revenue = -0.80(0)^2 + 20(0)
revenue = 0
From this calculation, we can see that the vertex of the revenue equation is at (150, 0). This means that the maximum revenue the caterer can make is $0, indicating that the caterer would not make any profit if 150 members attend the dinner.
To determine the upper limit for the number of members that can attend the dinner, we need to find the point where the caterer starts losing money. In this case, the caterer would start losing money when the revenue becomes negative.
Since the revenue equation is in the form of a parabola opening downward, the caterer would start losing money after the vertex. Therefore, the caterer should limit the number of attendees to 150 members to maximize revenue and avoid making a loss.