how many pairs of shoes can a dealer purchase for 8400$

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
buddy
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how many pairs of shoes can a dealer purchase for 8400$

Postby buddy » Sat Mar 21, 2009 1:07 pm

A company sells running shoes to dealers for $40 per pair if less than 50 pairs are ordered. If 50 or more pairs are ordered (up to 600), the price per pair is reduced at a rate of $0.04 times the number ordered. How many pairs can a dealer purchase for $8400?

i dont get it

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stapel_eliz
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Postby stapel_eliz » Sat Mar 21, 2009 3:32 pm

buddy wrote:A company sells running shoes to dealers for $40 per pair if less than 50 pairs are ordered. If 50 or more pairs are ordered (up to 600), the price per pair is reduced at a rate of $0.04 times the number ordered. How many pairs can a dealer purchase for $8400?

To find the pattern, start plugging values in, and see what happens. First, note that ($40/pr)(50 pr) = $2000, so clearly the dealer will be working with the "50 or more" case.

pairs |   price per   |        total        |
50 | 40 - 0.04(50) | (40 - 0.04(50))(50) |
51 | 40 - 0.04(51) | (40 - 0.04(51))(51) |
52 | 40 - 0.04(52) | (40 - 0.04(52))(52) |
53 | | |
... | | |
x | | |

Once you see the pattern, you can create the "cost" formula in terms of "x" pairs of shoes. Set your formula equal to "8400", and solve for the value of x.

If you get stuck, please reply showing all of your steps and reasoning so far. Thank you! :D

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Re: how many pairs of shoes can a dealer purchase for 8400$

Postby buddy » Sun Mar 22, 2009 6:45 pm

so is it (40 - 0.04(x))(x)? then wat?

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stapel_eliz
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Postby stapel_eliz » Mon Mar 23, 2009 3:16 pm

buddy wrote:so is it (40 - 0.04(x))(x)? then wat?

Now follow the rest of the instructions: Set the "cost" formula equal to the total given cost. Then solve the resulting quadratic equation for the number purchased.

:D

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Re: how many pairs of shoes can a dealer purchase for 8400$

Postby buddy » Sat Mar 28, 2009 9:47 pm

so (40 - 0.04(x))(x)=8400
but i get x=300 & x=700?? :confused:

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Postby stapel_eliz » Sun Mar 29, 2009 12:00 am

Re-read the original exercise. There are constraints included which will eliminate one of the solutions. :wink:

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Re: how many pairs of shoes can a dealer purchase for 8400$

Postby buddy » Sun Mar 29, 2009 7:07 pm

'up to 600' ok thnx


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