Sunday, February 6, 2011

The cost of producing x units of a product is given by C(x)=200+50x-50 ln x, x>=1. Find the minimum average cost.

The cost function is C(x) 200+50x-50lnx. x>=1


To find the minimum.


Solution:


We use the calculus criteria for minimum. A continuous derivable  function  f(x) has its minimum at  x= c if f'(c) = 0 and f '' (c)  is positive. This we apply to the cost function C(x):


C(x) = 200+50x-50lnx. Differentiating we get


C'(x) = (200+50x-50lnx)'


= (200)'+(50x)' - (50ln x)'


=0+50-1/x. Equate the C'(x) to zero and find x:


C'(x) = 0. gives 50-50/x = 0,  Or x = 50/50 = 1.


C''(1) = (C'(x))' at x= 1


=(50-50/x)' at x =1


=(50/x^2)at x=1


=50. So  at x=1,  C"(x) is positive. So C(x) is minimum for x = 1.


Therefore, the minimum average cost is C(1) = 200+50*1-50 ln 1


= 200+50-50*0 as ln1 =0


= 250

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