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by karthikpandian19 » Wed Jan 04, 2012 9:48 pm
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
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by ronnie1985 » Wed Jan 04, 2012 10:46 pm
The equation to maximize is 6x-0.03x^2
One can reduce the equation in perfect square form and get the answer as x=100 or use differential calculus to solve and get x = 100
(E) is answer
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by Anurag@Gurome » Wed Jan 04, 2012 10:54 pm
karthikpandian19 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
Price of a product, P = 6 - 0.03x implies x = (6 - P)/0.03 = 100(6 - P)/3
For maximum revenue, either the price should be maximum or the no. of products sold should be maximum.
Since x is the number of products sold, so it has to be a positive integer.
So, 6 - P should be a factor of 3
If P = 3.6, then x = 80; revenue = 3.6 * 80 = 288
If P = 3, then x = 100; revenue = 3 * 100 = 300
If P = 2.7, then x = 110; revenue = 2.7 * 110 = 297

Maximum possible revenue = 300

The correct answer is D.
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by karthikpandian19 » Thu Jan 05, 2012 9:43 pm
Is there any other method to solve this problem, without substituting values ?

Please explain
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by shankar.ashwin » Thu Jan 05, 2012 11:30 pm
Price of 1 article = 6 - 0.03x

Price of x articles = 6x - 0.03x^2

Since total price is = 6x - 0.03x^2, total price is maximum when 6x - 0.03x^2 is maximum.

We got to find maximum value of 'x' in 6x - 0.03x^2

As mentioned basic calculus is the easiest way to do it..

6 - 0.06x = 0

x = 100

WIthout calculus, the equation can be written as

- x^2 + 200x = Constant

-(x^2 - 200x + 100^2 - 100^2) = Constant

-(x-100)^2 + 100^2 = Constant

This attains a maximum when x = 100

when x =100, P = 6 - 0.03x = 3

Total revenue = xP = 300
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by karthikpandian19 » Fri Jan 06, 2012 4:06 am
I am so far from calculus now !!!!!
shankar.ashwin wrote:Price of 1 article = 6 - 0.03x

Price of x articles = 6x - 0.03x^2

Since total price is = 6x - 0.03x^2, total price is maximum when 6x - 0.03x^2 is maximum.

We got to find maximum value of 'x' in 6x - 0.03x^2

As mentioned basic calculus is the easiest way to do it..

6 - 0.06x = 0

x = 100

WIthout calculus, the equation can be written as

- x^2 + 200x = Constant

-(x^2 - 200x + 100^2 - 100^2) = Constant

-(x-100)^2 + 100^2 = Constant

This attains a maximum when x = 100

when x =100, P = 6 - 0.03x = 3

Total revenue = xP = 300
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by Abhishek009 » Fri Jan 06, 2012 9:37 am
karthikpandian19 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
In order to maximize P we must minimize x and he best option for such problems to my knowledge is checking out the least option given in answer choice...

Let's check option E.

P = 6 - 0.03 x

=> p = 6 - 0.03 ( 100 )
=> p = 6 - 3 => 2

Hence I will go for option E ..
Abhishek
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by GMATGuruNY » Fri Jan 06, 2012 10:41 am
karthikpandian19 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
ALWAYS LOOK AT THE ANSWER CHOICES.

Since the answer choices are all multiples of 100, the value of P (the selling price of each product) is almost certainly an integer, and the value of X (the number of products sold) is almost certainly a multiple of 100.

If x = 100, P = 6 - .03*100 = 3, and total revenue = 100*3 = 300.
If x = 200, P = 6 - .03*200 = 0, and total revenue = 200*0 = 0.
If x = 300, P = 6 - .03*300 = -3, and total revenue = 300(-3) = -900.
As x INCREASES, the total revenue DECREASES.
Thus, the maximum possible revenue = 300.

The correct answer is D.
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