The average cost of producing x units of a commodity is c(x)=21.4-0.0002x Find the marginal cost at a production level of 1000 units. In practical terms, what is the meaning of your answer?
Hmm I wasn't sure what marginal cost was, so I Googled it. "Marginal cost is the change in total cost when another unit is produced; average cost is the total cost divided by the number of goods produced." So marginal cost is the change at a specific moment, after a unit is sold. We can think of this as instantaneous change. So derivative, ya? :)\[\large\rm c'(1000)=?\]
sorry I've been helping someone
so c'(x)= .0002?
The one thing im confused about is there is no x left so??? What do you do with the 1000? Or can you put that in first?
\[\large\rm c(x)=21.4-0.0002x\]So\[\large\rm c'(x)=-0.0002\]right? with the negative? Evaluating at 1000 gives us,\[\large\rm c'(1000)=-0.0002\]
What is the meaning of this value? Hmmm thinking.
So that is just the answer? That just seemed so weird!
Omg someone posted something nasty on here :/
Well recall that polynomials always decrease by `one degree` when you take derivative. We have a straight line, (first degree x), which turns into a constant when differentiated. Constant means unchanging, no matter what x you plug in.
I'm not really sure how to interpret the meaning D: So the marginal cost to produce 1000 units is decreasing by 2 hundredths of a cent per unit produced. Something like that maybe? Bahhh I dunno.
hundredths?
This is a penny .01 This is a tenth of a penny .001 This is a hundredth of a penny .0001
ummmm?
I thought one hundredth is 2 after decimal
one hundredth of a whole unit is 2 after the decimal. But we're not saying that 1 is a penny we're saying that .01 is a penny
ohhhhhhhhhhhhhhhhhhh okayyyyyyy
so we have to move two places from here .01 ya? :o
yes :) this makes sense now wow I feel silly
:3
it doesn't really help that I haven't taken economics yet!!!
Same D:::
I'm not excited to take it I heard its kinda dumb
Well thank you for the help I might be on asking more questions later
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