When bicycles are sold for $300 each, a cycle store can sell 70 in a season. For every $25 increase in the price, the number sold drops by 10.
Represent the sales revenue as a function of the price.
The total sales revenue is $17,500. How many bicycles were sold? What is the price of one bicycle?
What range of prices will give us sales revenue that exceeds $18,000?
When bicycles are sold for $300 each, a cycle store can sell 70 in a season. For every $25 increase in the price, the number sold drops by 10.
Represent the sales revenue as a function of the price.
The total sales revenue is $17,500. How many bicycles were sold? What is the price of one bicycle?
What range of prices will give us sales revenue that exceeds $18,000?
The sales revenue (per bike) can be represented by ( 300 + 25x) where x represents the number of $25 increases
The revenue function, R(x) can be represented as
R(x) = ( 300 + 25x) ( 70 - 10x) = -250x^2 -1250x + 21000
For $17500, we have
17500 = -250x^2 - 1250x + 21000
250x^2 + 1250x - 3500 = 0 divide through by 10
25x^2 + 125x - 350 = 0 divide through by 25
x^2 + 5x - 14 = 0 factor
(x + 7) ( x - 2) = 0
Set each factor to 0 and solve for x .... when x = 2 ..... the price of one bike = (300 + 2(25)) = $350 and the number of bikes sold = (70 - 10(2) ) = 50
And when x = - 7 the price of a bike = (300 + 25(-7) ) = $125 and the number of bikes sold = (70 - 10(-7) ) = 140
To find the range of prices where sales revenue > $18000 look at the graph here :
https://www.desmos.com/calculator/gindgnqtxq
The graph shows that the revenue will > $18000 when x ≈ about -6 to x ≈ 1
In other words, when the price of a bike is from ≈ (300 + 25(-6) ) ≈ $ 150 to ≈ (300 + 25 (1) ) ≈ $325