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