Minimum average cost?



Why do you not show any work?

This web site is a cut above the answer mills.

You need to start showing some investment on your part.

 
If I'm reading this right, you are supposed to find a minimum of the C(x) function ???

The minimum (minima) will always occur where the first derivative is 0. But what's the first derivative of

C(x) = 4 x^2 + 100

? You wrote 8x + 100, which is incorrect. Start from here....

If I'm misunderstanding the problem, you're going to have to show more work and provide some explanations about why you're taking certain steps, but we're going to try to help the best we can.
 
chivox said:
If I'm reading this right, you are supposed to find a minimum of the C(x) function ???

The minimum (minima) will always occur where the first derivative is 0. But what's the first derivative of

C(x) = 4 x^2 + 100

? You wrote 8x + 100, which is incorrect. Start from here....
Click to expand...

Isn't 8x + 100 the derivative C(x)?
 
beachbunny619 said:
Isn't 8x + 100 the derivative C(x)?
Click to expand...

8x + 100 is the derivative of 4 x^2 + 100 x. Is that what C(x) is? I thought C(x) was 4 x^2 + 100 (the 100 being a constant term, which does not figure in the first derivative).

Anyway, once you have the first derivative, where is it equal to 0?
 
chivox said:
beachbunny619 said:
Isn't 8x + 100 the derivative C(x)?
Click to expand...

8x + 100 is the derivative of 4 x^2 + 100 x. Is that what C(x) is? I thought C(x) was 4 x^2 + 100 (the 100 being a constant term, which does not figure in the first derivative).

Anyway, once you have the first derivative, where is it equal to 0?
Click to expand...

Do I set the derivative = 0 or C(x)?
 
beachbunny619 said:
Do I set the derivative = 0 or C(x)?
Click to expand...

To find where C(x) is at a minimum, you need to know the x value where the first derivative equals 0. If it's suppose to be 8x+100, I think you already figured that out. I just thought that was a mistake, because when you originally wrote C(x), you wrote C(x) = 4 x^2 + 100.

So, let's just say you mistyped the C(x) function initially, and the first derivative is 8x + 100. Now, that equals 0, as you solved above,

8x + 100 = 0
8x = -100
x = -100/8

After you know the x value at which the minimum occurs, plug it back into C(x) to find the minimum cost. It's a two-step process. First, figure out where the minimum occurs by setting the first derivative to 0, and solve for x. Second, plug that x value back into the original function to find the (x, y) pair for that minimum on the graph.
 
chivox said:
beachbunny619 said:
Do I set the derivative = 0 or C(x)?
Click to expand...

To find where C(x) is at a minimum, you need to know the x value where the first derivative equals 0. If it's suppose to be 8x+100, I think you already figured that out. I just thought that was a mistake, because when you originally wrote C(x), you wrote C(x) = 4 x^2 + 100.

So, let's just say you mistyped the C(x) function initially, and the first derivative is 8x + 100. Now, that equals 0, as you solved above,

8x + 100 = 0
8x = -100
x = -100/8

After you know the x value at which the minimum occurs, plug it back into C(x) to find the minimum cost. It's a two-step process. First, figure out where the minimum occurs by setting the first derivative to 0, and solve for x. Second, plug that x value back into the original function to find the (x, y) pair for that minimum on the graph.
Click to expand...

I pluged -100/8 into x and got 725 but thats not right???
 
Well, if C(x) = 4 x^2 + 100, the derivative is NOT 8x + 100, and my work was just an example. It's all wrong.

Let's start at the beginning. What is C(x)??

And you might as well tell me, what's the answer supposed to be?
 
chivox said:
Well, if C(x) = 4 x^2 + 100, the derivative is NOT 8x + 100, and my work was just an example. It's all wrong.

Let's start at the beginning. What is C(x)??

And you might as well tell me, what's the answer supposed to be?
Click to expand...

OK C(x)=4x^2 + 100
 
Back up one more time, and then I think we've got it.

beachbunny619 said:
Therefore the first derivative is 8x.

At what x value does that equal 0?

so i got x=-8 --- nope. Solve for x: \(\displaystyle 8 x = 0\)
Click to expand...
 
chivox said:
Back up one more time, and then I think we've got it.

beachbunny619 said:
Therefore the first derivative is 8x.

At what x value does that equal 0?

so i got x=-8 --- nope. Solve for x: \(\displaystyle 8 x = 0\)
Click to expand...
Click to expand...

Wouldnt it just be 0
 
beachbunny619 said:
Wouldnt it just be 0
Click to expand...

And then 4 x^2 + 100 would be 100, right? Is that the answer? Because if it's not, then I have completely misunderstood what the problem was asking.

For sure, 4 x^2 + 100 is at a minimum at x = 0,
and the value of the function at x = 0 is 100.

Speaking as someone who has spent many years working with students, I have to encourage you to be honest with yourself. If this calculus class is too hard, you will not enjoy it after some of these initial lessons. And that would not lead to a good thing for you.

You have shown a little sloppiness in your algebra, like forgetting the negative sign up there with the 100/8, and in your calculus, like forgetting that constant terms just become 0 in the derivative, etc. I just don't want you to give up on math because it's too frustrating. It would be better to spend some time learning the algebra, trig, geometry, and those precalculus concepts before getting into calculus. If you're not solid in those, calculus is not the right place for you.

On the other hand, if you're just lazy with your math, that's your choice, I suppose. At least you have shown a very good initiative seeking help on this board, and that's a good start. I also hope your teacher is good, but the most important thing is for you to be honest with yourself. If you don't have the algebra stuff down cold, calculus isn't going to be fun after much longer. Maybe there's another way to brush up on your algebra, but it's going to come in handy for calculus.
 
chivox said:
… It would be better to spend some time learning the algebra, trig, geometry, and those precalculus concepts before getting into calculus. If you're not solid in those, calculus is not the right place for you …
Click to expand...


I told the beach bunny in two other threads that she's not ready to study calculus, but it appears that a moderator removed those comments.

I do not think that the beach bunny is ready to study calculus.

There.

I said it, again.

 
if total cost is ...

\(\displaystyle C(x) = 4x^2 + 100\) ,

then average cost per unit is ...

\(\displaystyle C_{avg} = 4x + \frac{100}{x}\)

this is the function to minimize.

don'tcha just luv business calc?
 
If s/he cannot find the minimum of C(x) - then I don't see a hope for the minimum of C[sub:2zvt2858]avg[/sub:2zvt2858].
 
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