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# PS

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Intern
Joined: 16 Feb 2012
Posts: 27
GPA: 3.57
Followers: 0

Kudos [?]: 55 [0], given: 0

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05 Jun 2012, 21:51
The price of a product manufactured at a company KTM is given by the following formula: P=6−0.03x , where P is the price of a single product, and x is the number of products sold. What is the maximum possible revenue for KTM?

a) 1000
b) 600
c)400
d)300
e)100

Easy method to solve this problem!!
Manager
Status: Rising GMAT Star
Joined: 05 Jun 2012
Posts: 133
Location: Philippines
Concentration: General Management, Finance
GMAT 1: 660 Q V
GPA: 3.22
WE: Corporate Finance (Consulting)
Followers: 7

Kudos [?]: 52 [0], given: 16

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05 Jun 2012, 22:10
rajman41 wrote:
The price of a product manufactured at a company KTM is given by the following formula: P=6−0.03x , where P is the price of a single product, and x is the number of products sold. What is the maximum possible revenue for KTM?

a) 1000
b) 600
c)400
d)300
e)100

Easy method to solve this problem!!

I solved this in less than 2 minutes using Calculus (derivatives).

First: know that the price is given by the formula p = 6 - 0.03x

Second: we know that revenue is given by the formula r = x * p

Now, we have:
revenue = price * x

revenue = (6 - 0.03x)x

revenue = 6x - 0.03x^2

Now let's get the first derivative

revenue (prime) = 6 - 0.06x

Equate the first derivative to 0 to determine the critical point

0 = 6 - 0.06x

0.06x = 6

x = 100 (this means that the revenue function is maximized when x = 100)

As such:

Revenue = (6 - 0.03(100))100

Revenue = 300

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Intern
Status: ISB 14...:)
Joined: 25 May 2012
Posts: 30
Location: India
Concentration: Strategy
Schools: ISB '14 (A)
GMAT 1: 750 Q51 V39
GPA: 3.62
WE: Engineering (Energy and Utilities)
Followers: 1

Kudos [?]: 34 [2] , given: 11

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05 Jun 2012, 23:43
2
KUDOS
The method suggested by gmatsaga is the best and foolproof one. But, here is an alternative one.

P varies from 6 to 0 in steps of 0.03(linearly)

X varies from 0 to 200 in steps of 1 (you get x=200 in P=0 case). This is also linear.

we need to maximize the revenue ie P.X which occurs at the midpoint.(As both P and X move linearly with opposite slopes)

So, the maximum point would be at P=3 and X=100. And the revenue will be 300.

PS: Sorry, if I have confused you....
Manager
Joined: 13 Feb 2012
Posts: 147
Location: Italy
Concentration: General Management, Entrepreneurship
GMAT 1: 560 Q36 V34
GPA: 3.1
WE: Sales (Transportation)
Followers: 4

Kudos [?]: 7 [1] , given: 85

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12 Jan 2013, 06:23
1
KUDOS
My two cents:

if they sell 10 products the price for each will be 5.70 since P=6-0.3, with a revenue of $57. If they sell 100 products, the price will drop to$3 since P=6-(0.03*100) = 3. The revenue in this case will be \$300.

If they sell anymore the price will be too low and the result will be less than 300.

Is my reasoning flawed?

This is how I solved it and got D.
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Andy

Director
Status: Gonna rock this time!!!
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Posts: 547
Location: India
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GMAT 2: 630 Q47 V29
WE: Information Technology (Computer Software)
Followers: 3

Kudos [?]: 61 [0], given: 562

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14 Feb 2013, 05:41
gmatsaga wrote:
rajman41 wrote:
The price of a product manufactured at a company KTM is given by the following formula: P=6−0.03x , where P is the price of a single product, and x is the number of products sold. What is the maximum possible revenue for KTM?

a) 1000
b) 600
c)400
d)300
e)100

Easy method to solve this problem!!

I solved this in less than 2 minutes using Calculus (derivatives).

First: know that the price is given by the formula p = 6 - 0.03x

Second: we know that revenue is given by the formula r = x * p

Now, we have:
revenue = price * x

revenue = (6 - 0.03x)x

revenue = 6x - 0.03x^2

Now let's get the first derivative

revenue (prime) = 6 - 0.06x

Equate the first derivative to 0 to determine the critical point

0 = 6 - 0.06x

0.06x = 6

x = 100 (this means that the revenue function is maximized when x = 100)

As such:

Revenue = (6 - 0.03(100))100

Revenue = 300

what on earth are derivatives
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Re: PS   [#permalink] 14 Feb 2013, 05:41
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