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Question 1108268: 2. 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 by Theo(13342)   (Show Source):
You can put this solution on YOUR website! number of members times revenue per member equals total revenue.
the current formula is therefore 200 * 30 = 6000
for every increase of 5 dollars in the membership fee, the number of members decreases by 20.
let x equal the the number of increases of 5 dollars in the revenue per member and the number of decreases of 20 in the number of members.
the formula becomes r = (200 - 20x) * (30 + 5x)
since the current revenue is 6000, and you want to know when the increase in the price will make the revenue go below 6000, then you get the following inequality.
6000 > (200 - 20x) * (30 + 5x)
this can also be written as (200 - 20x) * (30 + 5x) < 6000.
distribute the multiplication on the left side of this inequality to get:
6000 + 1000x - 600x - 100x^2 < 6000.
combine like terms to get 6000 + 400x - 100x^2 < 6000
subtract 6000 from both sides of this inequality to get 400x - 100x^2 < 0
divide both sides of this inequality by 100 to get 4x - x^2 < 0.
reorder the terms in descending order of degree to get -x^2 + 4x < 0.
to find the 0 points of this inequality, set -x^2 + 4x = 0 and solve for x.
factor out an x from the left side of the equation to get x * (-x + 4) = 0
x is either equal to 0, or -x + 4 is equal to 0.
solve for x and you will get x = 0 or x = 4.
your intervals will be x < 0, 0 < x < 4, x > 4
pick values in each of the intervals to test.
i chose x = -2, x = 2, x = 6
when x = -2, -x^2 + 4x = -4 - 8 = -12
when x = 2, -x^2 + 4x = -4 + 8 = 4
when x = 6, -x^2 + 4x = -36 + 24 = -12
-x^2 + 4x is less than 0 when x < 0.
-x^2 + 4x is greater than 0 when 0 < x < 4.
-x^2 + 4x is less than 0 when x > 4.
this says that your original equation will be less than 6000 when x is < 0 and when x > 4.
your original equation is r = (200 - 20x) * (30 + 5x)
set x = -2 and the equation becomes r = (200 - 20(-2)) * (30 + 5(-2)).
this results in r = 4800 which is less than 6000.
set x = 2 and the equation becomes r = (200 - 20(2)) * (30 + 5(2)).
this results in r = 6400 which is greater than 6000.
set x = 6 and the equation becomes r = (200 - 20(6)) * (30 + 5(6)).
this results in r = 4800 which is less than 6000.
you can graph the equation of r = (200 - 20x) * (30 + 5x) by setting r equal to y and graphing y = (200 - 20x) * (30 + 5x).
the graph will look like this:
you can see from the graph that, when x is < 0 or > 4, that y is less than 6000.
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