Business Calculus 1: Average Cost Function & Min. Production

[Hannelore]

1. C(x)=2200 + 5x + 0.002x^2 + 0.00013x^3

(a) Find the average cost function and sketch it

(b.) At what production level is the average cost minimized?

Homework Equations

My book defines the average cost as c(x)=C(x)/x

And the average cost is minimized when C'(x) = c(x)

The Attempt at a Solution

I'm supposed to be teaching the class this and the teacher gave me the answers for this problem but I don't know how he got them.

He has the final answer of (a) at c(x) = 2200/x + 0.002x + 0.00013x^2
But I have it at c(x) = (2200 + 5x + 0.0002x^2 + 0.00013x^3)/x
I thought that he might have taken the derivative however the numbers aren't right in his answer for that to have happened.

 

[hotcommodity]

Your answer for part a, and your teachers answer for part a are the same. The only difference is that he simplified his function by canceling out the x variables in each respective term.

If I have C(x) = 10x^{2} + x and I want to divide that function by x, I have c(x) = \frac{10x^{2}}{x} + \frac{x}{x} which simplifies to c(x) = 10x + 1.

As for part b, you must take the derivative to find the minimum. Set the derivative equal to zero, find what value x must be for the function to equal zero (this is called the critical point, and you may have more than one), and plug that x value, or values, into the original equation, and see which value yields a minimum.

 

[Hannelore]

I'm going to need more help with part b.

I took the derivative of C(x) which is C'(x) = 5 + .004x + .00039x^2

I set it to zero and then I factored it but I got unreal numbers, negatives under the square root sign.

So, I did it the way the book instructed and I set C'(x)=c(x) and this is what I got:

.0000000507x^4 + .0000013x^3 + .006998x^2 + 5.01x = 2175

...and I don't know what to do next

 

[HallsofIvy]

You weren't asked to find minimum cost C(x) (you are right, the derivative of C has only non-real zeros so there is no local minimum). You were asked to find the minimum average cost, c(x).

As we have already worked out, c(x) = 2200/x + 0.002x + 0.00013x^2 which is the same as c(x) = (2200 + 5x + 0.0002x^2 + 0.00013x^3)/x= 2200/x+ 5x/x+ 0.0002x^2/x+ 0.00013x^3/x= 2200 x^-1+ 0.002x+ 0.00013x². Differentiate that.

(The derivative of 1/x= x^-1 is, using either the quotient rule on 1/x or the power rule on x^-1, -x^-2= -1/x²)