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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?

Micheala95  May 13, 2017
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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

CPhill  May 14, 2017

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