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# A salesperson sells shoes. At a price of \$32, he can sell 80

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Joined: 04 Jan 2008
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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

Kudos [?]: 104 [0], given: 0

SVP
Joined: 07 Nov 2007
Posts: 1792

Kudos [?]: 1061 [0], given: 5

Location: New York

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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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Kudos [?]: 1061 [0], given: 5

VP
Joined: 17 Jun 2008
Posts: 1374

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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
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Its Now Or Never

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Joined: 17 Jun 2008
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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?

Kudos [?]: 279 [0], given: 0

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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
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Joined: 05 Jul 2006
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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
\$20

the premisis of the question is specefying ( for every 2 dollars discount ) thus the answer has to be even.( 32 - 24 - 20)

however 32 is obviously ruled out

then it is either ( 24 - 20) just substitute and compare

the first 80 shoes are not relevent thus we need to calculate the excess in rev from additional sales due to discount. ie:

- 24*(20*4) = 1920

- 20 *(20*5) = 2000

Kudos [?]: 430 [0], given: 49

Re: Zumit PS 014   [#permalink] 16 Sep 2008, 13:43
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