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rajman41
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PS [#permalink] New post   05 Jun 2012, 22:51
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!!

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Re: PS [#permalink] New post   05 Jun 2012, 23:10
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

Answer (D)
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Re: PS [#permalink] New post   06 Jun 2012, 00:43
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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.... :-D
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Re: PS [#permalink] New post   12 Jan 2013, 07:23
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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Re: PS [#permalink] New post   14 Feb 2013, 06:41
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

Answer (D)


what on earth are derivatives :shock: :shock: :shock: :shock: :shock: :shock:

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Re: PS [#permalink]
14 Feb 2013, 06:41
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