how would i do this quadratic word problem

A local daycare centre charges $65 per day to care for a child. The daycare currently cares for 150 children per day. A survey shows that the enrollment in the daycare will increase by 10 children for each $5 decrease in daily fee, and would decrease similarly if the daily fees are increased.

What daily fee would result in the greatest revenue for the daycare centre?

Let the number of $5 decreases be n

so cost per child = 65 - 5n
number of children = 150 + 10n

Revenue = (65-5n)(150+10n)
= 9750 - 100n - 50n^2
d(Revenue)/dx = -100 -100n = 0 for a max/min of Revenue

100n = -100
n = -1
so the number of decreases is -1, or there should be an increase of
$5 for a daily fee of $70

check:
at $65 , number is 150, R = 9750
at $60, number is 160, R = 9600
at $70, number is 140, R = 9800
at $75, number is 130, R = 9750

x = amount of increase of fee

enrollment = n = 150 - 10 (x/5)
= 150 - 2x
revenue = n * (65+x)
so
revenue = (150-2x)(65+x)
R = 9750 + 20x -2 x^2
find the vertex
R/2 = 4875 +10 x -x^2
x^2 -10 x = -(R/2) + 4875
x^2 - 10 x + 100 = -(R/2) +4975
(x-10)^2 =
so vertex at x = 10
so fee at vertex = 65+10 = 75

To solve this quadratic word problem and find the daily fee that would result in the greatest revenue for the daycare center, we need to follow these steps:

Step 1: Determine the revenue equation
Let's start by defining some variables:
- x: the number of $5 decreases in the daily fee.
- y: the number of children enrolled in the daycare center.
- f: the initial daily fee of $65.

The revenue generated by the daycare center can be calculated by multiplying the number of children (y) with the daily fee (f - 5x):
Revenue = (f - 5x)y

Step 2: Determine the relationship between x and y
According to the problem, the survey shows that for every $5 decrease in the daily fee, the enrollment in the daycare center will increase by 10 children, and it will decrease similarly if the daily fees are increased. This means the number of children enrolled (y) is dependent on the number of $5 decreases in the daily fee (x):

y = 150 + 10x

Step 3: Substitute the value of y into the revenue equation
Now we can substitute the value of y from the previous step into the revenue equation:

Revenue = (f - 5x)(150 + 10x)

Step 4: Expand and simplify the revenue equation
To find the maximum revenue, we need to find the value of x that will maximize the revenue equation. Expand and simplify the equation:

Revenue = 150f + 300x - 50x^2

Step 5: Find the vertex of the parabolic equation
We have now transformed the revenue equation into a quadratic equation in the form of y = ax^2 + bx + c, where a = -50, b = 300, and c = 150f. The maximum revenue is achieved at the vertex of the parabolic equation.

The x-coordinate of the vertex can be determined using the formula:
x = -b / (2a)

In our case, x = -300 / (2 * -50)

Step 6: Calculate the optimal daily fee (f) for maximum revenue
Now that we have the x-coordinate of the vertex, we can substitute it back into the equation y = 150 + 10x to find out the corresponding number of children enrolled.

Once we have the number of children enrolled, we can calculate the optimal daily fee (f) by multiplying it with $65 (the initial daily fee):

f = (150 + 10x) * 65

Substitute the value of x we calculated in the previous step and solve for f.

Finally, the resulting value of f will be the daily fee that would result in the greatest revenue for the daycare center.