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

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The cost function for producing x units is defined as f(x)=800+110x-110ln(x), and the goal is to find the minimum average cost for x >= 1. The average cost is calculated as C(x)=(800+110x-110ln(x))/x, and the marginal cost is determined by the derivative C'(x)=(110ln(x)-910)/x^2. Setting C'(x) to zero leads to the critical point x=e^(910/110), but this value is reported as incorrect by Webwork. The second derivative C''(x)=(1930-220ln(x))/x^3 indicates an increasing function, suggesting further analysis is needed to confirm the minimum average cost.
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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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