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05 Jun 2012, 22:51
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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!!
a) 1000
b) 600
c)400
d)300
e)100
Easy method to solve this problem!!
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05 Jun 2012, 23:10
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rajman41
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)
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....
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....
