Finding minimum average cost for f(x)=800+110x-110ln(x)

[aft_lizard01]

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.

 

[HallsofIvy]

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.