# Marginal Cost

• November 6th 2007, 08:10 PM
Truthbetold
Marginal Cost
Suppose that the dollar cost of producing x washing machines is $c(x)= 2000 + 100x -.1x^2$

Simplified it is: $c(x)= x^2 - 1000x -20000$

(a) Find the average cost of producing 100 washing machines.

Don't get it or what to use.

(b) Find the marginal cost when 100 machines are produced.
I know how to do this one, Taek the derivative and plug in x.

(c) Show the marginal cost when 100 washing machines are produced is approximately the cost of producing one more washing machine after the first 100 have been made, by calculating the latter cost directly.

Don't understand what they want.

For the average cost, I think you take the derivative of it and then plug x= 100.

However, I got a negative number.

In the book, it says " [the difference quotient without the limit] = the average cost of each of the additional h tons produced.

So, do I plug it in and just deal with the highly messy equation or am I missing something?
• November 7th 2007, 10:35 PM
Truthbetold
Bump. Fixed LaTex.

Still don't get it.
• November 7th 2007, 10:47 PM
topsquark
Quote:

Originally Posted by Truthbetold
Suppose that the dollar cost of producing x washing machines is $c(x)= 2000 + 100x -.1x^2$

Simplified it is: $c(x)= x^2 - 1000x -20000$

(a) Find the average cost of producing 100 washing machines.

Don't get it or what to use.

(b) Find the marginal cost when 100 machines are produced.
I know how to do this one, Taek the derivative and plug in x.

(c) Show the marginal cost when 100 washing machines are produced is approximately the cost of producing one more washing machine after the first 100 have been made, by calculating the latter cost directly.

Don't understand what they want.

For the average cost, I think you take the derivative of it and then plug x= 100.

However, I got a negative number.

In the book, it says " [the difference quotient without the limit] = the average cost of each of the additional h tons produced.

So, do I plug it in and just deal with the highly messy equation or am I missing something?

You know, when you "simplified" your cost function you created a problem where you need to do the calculations, negate the answer, and then divide it by 10. It's simpler to work with the given function.

a) Average cost to make the first 100 machines:
$c_{ave} = \frac{c(100) - c(0)}{100 - 0} = \frac{2000 + 100 \cdot 100 - 0.1 \cdot 100^2 - 2000}{100} = 90$

b) You said you can do this, so I won't.

c) You have your answer to b). Now show that it is approximately equal to the cost of making 100 + 1 machines. (That is, the cost of making 100 machines, then starting all over to make 1 machine, so this is not c(101). )

-Dan
• November 8th 2007, 07:16 PM
Truthbetold
Quote:

Originally Posted by topsquark
You know, when you "simplified" your cost function you created a problem where you need to do the calculations, negate the answer, and then divide it by 10. It's simpler to work with the given function.

a) Average cost to make the first 100 machines:
$c_{ave} = \frac{c(100) - c(0)}{100 - 0} = \frac{2000 + 100 \cdot 100 - 0.1 \cdot 100^2 - 2000}{100} = 90$

-Dan

The answer, according to the book, is actually 110.
I have not done it yet, but I assume you made a computation error because the formula makes sense.
• November 11th 2007, 07:11 PM
Truthbetold
Going over it again, I don't get C.

$\frac{c(1)-c(0)}{1-0}$

Gets something like 99.9, which is 20 off. Answer is 79.9.
I think I misunderstood.

I tried a couple of things.
Ex. plugging 1 into original equation, making C(100) + c(1) - c(0 (and even with another c(0)).

So I'm lost.

Thanks!