Find Relative Extrema: Min Avg Cost - f(x) & x>=1

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SUMMARY

The discussion focuses on finding the minimum average cost of producing x units, represented by the function f(x) = 800 + 110x - 110ln(x). The average cost function is derived as C(x) = (800 + 110x - 110ln(x)) / x. The first derivative, C'(x) = (110ln(x) - 910) / x^2, is set to zero to find critical points, leading to the incorrect solution x = e^(910/110). The second derivative, C''(x) = (1930 - 220ln(x)) / x^3, indicates the nature of the critical points, but the initial calculation of x is confirmed incorrect by Webwork.

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Homework Statement



The cost of producing x units is given by:

f(x)=800+110x-110ln(x)

Find the minimum average cost. x >= 1



Homework Equations



Cost=f(x)/x

Marginal cost is C'(x)


The Attempt at a Solution



This is what I have done so far:

C(x)=(800+110x-110ln(x))/x

C'(x)= (110ln(x)-910)/x^2

setting C'(x)=0

I get x=e^(910/110)

C''(x)=(1930-220ln(x))/x^3

Plugging in my value for x I get an increasing number validating, or I thought my x=value. Webwork tells me that e^(910/110) is incorrect for the minimum cost.
 
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aft_lizard01 said:

Homework Statement



The cost of producing x units is given by:

f(x)=800+110x-110ln(x)

Find the minimum average cost. x >= 1



Homework Equations



Cost=f(x)/x

Marginal cost is C'(x)


The Attempt at a Solution



This is what I have done so far:

C(x)=(800+110x-110ln(x))/x

C'(x)= (110ln(x)-910)/x^2
This derivative is incorrect. Please show how you got it, step by step.


setting C'(x)=0

I get x=e^(910/110)

C''(x)=(1930-220ln(x))/x^3

Plugging in my value for x I get an increasing number validating, or I thought my x=value. Webwork tells me that e^(910/110) is incorrect for the minimum cost.
 

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