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Category > Calculus Posted 10 Jul 2017 My Price 8.00

The cost C=16x2+6400 

The cost C=16x2+6400 

Find the (x,y) coordinates of the average cost function when the slope of the tangent line is horizontal. 

 

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. 

 

AvgCost C = (16x2+6400)/x = 16x + (6400/x) 

 

Average Cost C' = 

N = 16x2+6400 ----> N'= 32x 

D = x ------------------> D' = 1 

C' = (x)(32x) - (16x2+6400)(1) 

Average Cost C' = 16 - (6400/x2) 

 

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? 

 

16 - (6400/x2) = 0 

16 = (6400/x2) 

16x2 = 6400 

x^2 = 400 

x = 200 

 

Does it mean the point is (x,y) = (200, 0)? Or I got this whole thing wrong.

Answers

(15)
Status NEW Posted 10 Jul 2017 01:07 AM My Price 8.00

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