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# A salesperson sells shoes. At a price of $32, he can sell 80  post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message Manager Joined: 04 Jan 2008 Posts: 119 Followers: 3 Kudos [?]: 81 [0], given: 0 A salesperson sells shoes. At a price of$32, he can sell 80 [#permalink]

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12 Sep 2008, 06:43
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A salesperson sells shoes. At a price of $32, he can sell 80 pairs of shoes. For every 2 dollars discount on the price of the shoes, 20 more pairs of shoes are sold. At what price will the salesperson have the highest revenue from shoe sales.$32
$27$24
$21$20
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12 Sep 2008, 21:40
dancinggeometry wrote:
A salesperson sells shoes. At a price of $32, he can sell 80 pairs of shoes. For every 2 dollars discount on the price of the shoes, 20 more pairs of shoes are sold. At what price will the salesperson have the highest revenue from shoe sales.$32
$27$24
$21$20

What is "ZUMIT?"

Is it zoom it?
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Joined: 17 Jun 2008
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14 Sep 2008, 03:28
dancinggeometry wrote:
A salesperson sells shoes. At a price of $32, he can sell 80 pairs of shoes. For every 2 dollars discount on the price of the shoes, 20 more pairs of shoes are sold. At what price will the salesperson have the highest revenue from shoe sales.$32
$27$24
$21$20

IMO E
32*80 = revenue in norml scenario
acc to this ques ,say we decrease the shoe sales n times by 2 dollars,20n additional pair of shoes sold ,hence new revenue is
(32-2n)(80+20n)=R
R is max when n=6 hence 32-2*6=20 $is the answer _________________ cheers Its Now Or Never SVP Joined: 17 Jun 2008 Posts: 1553 Followers: 11 Kudos [?]: 264 [0], given: 0 Re: Zumit PS 014 [#permalink] ### Show Tags 15 Sep 2008, 00:03 spriya wrote: IMO E 32*80 = revenue in norml scenario acc to this ques ,say we decrease the shoe sales n times by 2 dollars,20n additional pair of shoes sold ,hence new revenue is (32-2n)(80+20n)=R R is max when n=6 hence 32-2*6=20$ is the answer

Good explanation. But, how did you arrive at n = 6 for R to be the max?
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15 Sep 2008, 01:02
scthakur wrote:
spriya wrote:
IMO E
32*80 = revenue in norml scenario
acc to this ques ,say we decrease the shoe sales n times by 2 dollars,20n additional pair of shoes sold ,hence new revenue is
(32-2n)(80+20n)=R
R is max when n=6 hence 32-2*6=20 $is the answer Good explanation. But, how did you arrive at n = 6 for R to be the max? (32-2n)(80+20n)=R = 2(16-n)*20*(4+n) = 80(-n^2+12n+20) f(n)=(-n^2+12n+20) derivative of f(n)=f'(n)=-2n+12=0 ( n is max or min) n=6 _________________ Your attitude determines your altitude Smiling wins more friends than frowning SVP Joined: 05 Jul 2006 Posts: 1747 Followers: 6 Kudos [?]: 358 [0], given: 49 Re: Zumit PS 014 [#permalink] ### Show Tags 16 Sep 2008, 13:43 dancinggeometry wrote: A salesperson sells shoes. At a price of$32, he can sell 80 pairs of shoes. For every 2 dollars discount on the price of the shoes, 20 more pairs of shoes are sold. At what price will the salesperson have the highest revenue from shoe sales.

$32$27
$24$21