What kind of problem is this?

Question

The gym charges $180 for a yearly membership. There are currently 1000 members. For every $5 increase, the gym will lose 10 members. How much should the gym charge to maximuze its revenue?

 

What type of problem is this: linear,quadratic,polynomial,radical,exponential,or rational?

Answer 1

Let x be the number of increases in $5 increments, then

180 + 5x is the yearly fee associated with the x increments.

1000 - 10x is the number of members associated with the x increments.

Revenue = Fee x Number of members

R(x) = (180 + 5x)(1000 - 10x) = -50(x+36)(x-100) = -50(x² - 64x - 3600) = -50(x-32)² + 231,200

 

This is the quadratic equation of a parabola which opens downward, so the maximum revenue is at the vertex of R(x).

 

The new fee is 180 +5(32) = $340.

 

Note:

The new revenue is 231,200.  Compare this with the revenue for x = 0, you should get the current revenue of $180,000.

 

 

 

Answer 2

x represents an increase

r = (180 + 5x)(1000 - 10x)

 

This is a quadratic function.