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?
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)
Revenue = (6 - 0.03(100))100
Revenue = 300
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