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May 7, 2010

Profit Maximization

A manufacturer can sell X items (X >= 0) at a price of Rs. (330 - X)
each. The cost of producing X items is ((X^2) + 10X + 12).
Determine the no. of items sold so that the manufacturer can make
the maximum profit..


João Marques said...

my answer is 80.
we have the formula of the income money for x items:
and we have the formula for the cost:

we want to maximize the profit
max (income - cost)

= max(-2*x^2+320*x-12)

parabola.... max point is x=-b/2*a
=-320/-4 = 80

Pratap said...

I second what Marques answer is.


Profit = (330-x)x - (x^2 +10x+12)
= 320x - 2x^2-12
= 2x(160-x) -12

So we need to find max (2x(160-x))
which is achieved at 80

Ali said...

yes that is true, the maximum point is when the derivative of -2x^2+320x-12 equals 0 (horizontal slope)

that is -4x+320=0 ==> x=-320/-4 = 80