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MBA, Ph.D in Management
Harvard university
Feb-1997 - Aug-2003
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Strayer University
Jan-2007 - Present
The cost C=16x2+6400Â
Find the (x,y) coordinates of the average cost function when the slope of the tangent line is horizontal.Â
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Do I just find the derivative of the cost formula ? Or do I have to find the AvgCost forumla first and then calculate its' derivative and use that formula? I did the latter since I wasn't sure.Â
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AvgCost C = (16x2+6400)/x = 16x + (6400/x)Â
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Average Cost C' =Â
N = 16x2+6400 ----> N'= 32xÂ
D = x ------------------> D' = 1Â
C' = (x)(32x) - (16x2+6400)(1)Â
Average Cost C' = 16 - (6400/x2)Â
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From that point I don't know where to find (x,y) coordinates. I think I saw somewhere you set x=0 with the C' formula... so here goes?Â
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16 - (6400/x2) = 0Â
16 = (6400/x2)Â
16x2Â = 6400Â
x^2 = 400Â
x = 200Â
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Does it mean the point is (x,y) = (200, 0)? Or I got this whole thing wrong.
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