Question

1. A gym has 200 members who pay $30 per month for unlimited use of the gym’s equipment. A survey of the members indicates that for each $5 increase in the monthly fee, the gym will lose 20 members. This means that the revenue R from fees, which is currently $6000 per month, will become , where f is a whole number of $5 fee increases. Write and solve a quadratic inequality to answer the questions: For what numbers of $5 fee increases will the revenue from fees actually be less than its current value?

Answer 1

200 members pay $30 each
Current revenue = 200 x $30
= $6000

For each $5 increase in  fee, 20 members will leave
For i number of $5 increase, 20i members will leave
New number of members = 200 - 20i

For i number of $5 increase, new fee = 30 + 5i

Therefore, new revenue = (200 - 20i) x (30 + 5i)
= 200(30 + 5i) - 20f(30 + 5i)
= 6000 + 1000i - 600i - 100i²
= 6000 + 400i - 100i²

In the form of a quadratic equation, 
-100i² + 400i + 6000
R(i) = -100i² + 400i + 6000

This is not our answer yet.
We are told to write a quadratic inequality for what numbers of $5 fee increases will the revenue from fees actually be less than its current value

Current value = 6000

Therefore, we are looking for i, when R(i) < 6000
Remember, R(i) = -100i² + 400i + 6000

Therefore,
R(i) < 6000 = -100i² + 400i + 6000 < 6000

We are to solve the quadratic inequality,
-100i² + 400i + 6000 < 6000

⇒ -100i² + 400i < 0 (Subtract 6000 from both sides)
⇒ 400i < 100i²  (Add 100i to both sides)
⇒ 4i < i² (Divide both sides by 100)
⇒ 4 < i (Divide both sides by i)

4 < i can be rewritten as i > 4

Therefore, for more than 4 numbers of $5 increase in the fees, the revenue from fees will actually be less than its current value.