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..
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..
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3 comments:
my answer is 80.
we have the formula of the income money for x items:
x*(330-x)
and we have the formula for the cost:
x^2+10*x+12
we want to maximize the profit
max (income - cost)
max(330*x-x^2-x^2-10*x-12)
= max(-2*x^2+320*x-12)
parabola.... max point is x=-b/2*a
=-320/-4 = 80
I second what Marques answer is.
80
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
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
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