Please help!! =D a) At what level of production will the marginal cost be minimized? b) At what level of production will the average cost A(x)=C(x)/x be minimized? C(x)= x^(3)-24x^(2)+192x+338
In both instances, you're looking for critical points. Do you know how to find these?
I have an example and not sure how they got the answer.
Well, in part a your answer will be based on the deriviative of C(x). If you're not familiar with derivatives yet, then I can give you some rules.
Yes, I can find the first part which is 8. And I get through the second part until the equation. You want me to show which part.?
Yeah
f(x)= 2x^(3)-12x^(2)-338
If the average cost is C(x)/x, shouldn't you divide C(x) by x?
Yep! That is the part that i confused on. Hold on. let me go back
So, x inn this case will be 8?
Not sure!! can you explain that to me? please!
8 is the answer for part a, but if I'm right about the equation for b, then you need to set the derivative of that equation equal to zero, just as you did in a
I did everything and i got 6. but the answer was 13. Still don't get it...
@jim_thompson5910 help me if you are not busy, please. :)
hmm not sure on this one, seems like something is missing
Really?
I graphed C(x)/x and got a hyperbola...which means that it has no minimum, so there has to be restrictions somewhere
Ohh waoo!!
Can we try another one?
sure
C(x)= x^(3)-30x^(2)+300x+512 Thanks!
That one already start it and i got 10. which is correct but the second part is where i am lost.
Not sure what you mean when you say "i got 10". How did you get it? What exactly is it asking you to do here?
a) At what level of production will the marginal cost be minimized? I started yesterday
C(x)= x^(3)-30x^(2)+300x+512 is the cost function, so the marginal cost function is C'(x) = 3x^2 - 60x + 300
The min of that function will occur at the vertex
3(x-10)^2=0
good
so the vertex is at (10,0) which means that the min is y = 0
Ok!
How do i find it the b) part??.
i'm not 100% sure...i'm getting another hyperbola when I graph C(x)/x
No idea on what I am doing, why the answer is 16?
ok.
There must be a domain somewhere...that's my guess
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