# Thread: Cost Function question [Derivatives]

1. ## Cost Function question [Derivatives]

The cost function for the manufacture of mp3 players is $\displaystyle C(x) = 25000 + 20x + 0.001x^2$ Dollars.

a) How many mp3 players should be manufactures to minimize the cost?
b) What is the resulting average cost of an mp3 player?

Note: X= number of mp3 manufactures

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I'm stumped. This question was given to me by my Finite Maths teacher under the topic of 'Application of Derivatives', and I don't seem to be making any progress. I'd appreciate it if you guys could help me out

thanks!

EDIT: Correction in the problem

2. Originally Posted by Shamanistic
The cost function for the manufacture of mp3 players is $\displaystyle C(x) = 25000 + 20x + 0.001x^2$ Dollars.

a) How many mp3 players should be manufactures to minimize the cost?
b) What is the resulting average cost of an mp3 player?

Note: X= number of mp3 manufactures

====

I'm stumped. This question was given to me by my Finite Maths teacher under the topic of 'Application of Derivatives', and I don't seem to be making any progress. I'd appreciate it if you guys could help me out

thanks!

EDIT: Correction in the problem
For A, if im correct u need the derivation of the formula and setting that equal to zero.

For B, u need to divide it by X, since costs divided production gives the average costs per product.

3. Originally Posted by Massachusetts Cowboys
For A, if im correct u need the derivation of the formula and setting that equal to zero.

For B, u need to divide it by X, since costs divided production gives the average costs per product.
Yes, this is completely correct. Take the derivative of $\displaystyle C(x) = 25000 + 20x + 0.001x^2$ and set it equal to 0. Solve that equation for x.
When Massachusetts Cowboy say "divide it by x", he is referring to C(x) for the value of x you found in (a). Find the cost of manufacturing that number of mpe players and divide by that value of x.

4. A friend of mine told me to do this:

The cost function tells you the overall cost of making the units. What you need is the cost per unit function, which is (25000+20x+0.001x^2)/x.

That has a derivative (20 + 0.002 x)/x - (25000 + 20 x + 0.001 x^2)/x^2 which has solutions at -5000 and 5000.

So to minimise costs, manufacture 5000. And if you plug 5000 back into the equation then the overall cost is 25000 + (20 * 5000) + (0.001*(5000^2)) = 25000 + 100000 + 25000 = 150000, or cost per unit of 30.

What do you guys think?