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TL DR
Registered User regular

Hi, H/A. As usual, I will be taking full advantage of our resident math gurus in my time of need. Thanks in advance!

Second problem.

That gives me**C(q) = q^3 - 90q + 3500q**.

Marginal cost would be the derivative, correct?**C'(q) = 3q^2 - 180q + 3500**

I believe the next step is to set the function equal to zero to get the critical point(s).

**0 = 3q^2 - 180q + 3500**

**3500 = 3q^2 - 180q**

And here I am stuck. Thanks again for the assistance.

Spoiler:

Second problem.

So to get C(q) I would just multiply the a(q) function by q, since (average cost per unit) times (number of units) = (total cost), right?The average cost per item to produce q items is given bya(q) = q^2 - 90q +3500. Find the minimum value of the marginal cost. (Hint: First find the total cost C(q), and then MC(q). Then minimize MC(q).)

That gives me

Marginal cost would be the derivative, correct?

I believe the next step is to set the function equal to zero to get the critical point(s).

And here I am stuck. Thanks again for the assistance.

## Posts

EDIT: Ahh, thought you meant the number 0.994 itself. I missed the ln() bit.

The derivative of "a^x" is "(a^x)[ln(x)]." Most people just memorize this, but the reason is because of the chain rule: "a^x" is actually "e^[x ln(a)]." That is where ln(0.994) comes from.

Yeah. 0.994 represents the change in population. However, if you just take P(10) = 5.6 (0.994)^10, you end up with about 5.273 . Compared to the original 5.6, that's a difference of -0.327 .

The answer I need to be arriving at is -0.031733

So the chain rule as I understand it is (derivative of outside function) (derivative of inside function) or f'(g(x)) times g'(x).

So I'd get something like

P(t) = 5.6(0.994)^t

P'(t) = 5.6 (0.994)^t ln(0.994)

Where ln(0.994) is the derivative of the ( )^10 function? Then substitute 10 and get

P'(10) = 5.6 (0.994)^10 ln(0.994)

= 5.24

???

Are you certain?

It seems to me that since a^x = e^[x ln(a)], it's far more likely that ln(a) will appear in the result than ln(x).

In fact, Wikipedia seems to think that the derivative is (a^x)ln(a), for a > 0.

Edit: Here we go:

(e^[ln(a) x]) ' = [ x ln(a) ] ' * e^[ln(a) x]

=> (a^x) ' = ln(a) * e^[ln(a) x] = ln(a) * (a^x)

It's possible I copied something down wrong, and I don't see how I got from P'(10) to my answer. We do have a copy of the practice test with answers though, and it supports the -0.031733 number.

P0 is the initial population at time zero (year 2000)

e is Euler's number (roughly 2.71, use e on your calculator for best results)

r is the rate (generally the unknown)

t is the time (10 years, referring to the period between 2000 and 2010)

Can you phrase the question this way? Otherwise I'm at a loss as to help you. I've only seen this type of question phrased like this.

Here's how I would do it (I know this is repeating and I think you know the right answer now, but I'll just write it out in hopes it solidifies it):

P(t) = 5.6 * (0.994)^t

= 5.6 * e^(t ln(0.994))

P'(t) = (5.6 * e^(t ln(0.994)))'

let f(t) = t * ln(0.994)

f'(t) = ln(0.994)

So P'(t) = (5.6 * e^f(t))'

= 5.6 * (e^f(t))' (since we can take constants out of derivatives)

= 5.6 * f'(t) * e^f(t) (by chain rule)

= 5.6 * ln(0.994) * e^(t ln(0.994))

= 5.6 * ln(0.994) * (0.994)^t

So P'(10) = 5.6 * ln(0.994) * (0.994)^10

This is correct.

You must have typed something into your calculator wrong to end up with 5.24. If you evaluate that correctly, you get -0.0317328642 (thanks google calculator). That's in millions of ppl/year so you need to multiply by a million to get it into people/year.

So to get C(q) I would just multiply the a(q) function by q, since (average cost per unit) times (number of units) = (total cost), right?

That gives me

C(q) = q^3 - 90q + 3500q.Marginal cost would be the derivative, correct?

C'(q) = 3q^2 - 180q + 3500I believe the next step is to set the function equal to zero to get the critical point(s).

0 = 3q^2 - 180q + 35003500 = 3q^2 - 180qAnd here I am stuck. Thanks again for the assistance.

C''(q) = 6q - 180q = 3

Then what, evaluate C'(3)?

Yes. Q=30 is the quantity, so to find the price plug back into your Marginal Cost function.

Hopefully you understand the general concept of minimizing functions (it seems like you mostly do but were a little confused here). If not, say so. But, to minimize C, you set C'=0. To minimize C', you set C''=0, etc.

EDIT: Q is actually 30. Think you just typed it wrong.

Yeah, twas just a typo. I got it. Thanks!

Next problem:

Minimize the cost of this fence. The area is 200 square feet, and the fencing along 3 sides costs $1/foot. The third side is $3/foot.

I have:

Area = 200

XY= 200

C = $1 (2x+y) + $3 (y)

C = 2x+4y

And we have to minimize Y.

So I guess something like

200/y = x

C = 2x + 4(800/x)

C = 2x + 3200/x

-3200 = 2x^2

-1600 = x^2

... I think I'm missing something. Do I want to make 3200/x into 3200 * x^-1 ?

You messed up. First of all, C = 2x + (800/x) NOT 3200/x.

Find C'. Set C'=0. Solve for x. Then plug X into xy=200 to solve for y.

Thanks for the help. Wish me luck.

edit: Oh, looks like the exam is actually at 4:30. Gonna review some more

Can I evaluate this as f(1) and then derivate that, or should I find the derivative and then plug in 1 for x?

You have to find the derivative and then plug in. If you evaluated f(1) you would get a constant, and the derivative of that would be 0.

Hmm. I'm looking at something like

[2(3x-2) - 3(2x+1)] / (3x-2)^2

[(6x-4) - (6x+3] / (3x-2)^2 for the inside function via quotient rule. Correct? Then continue by taking (derivative inside function) (outside function) + (inside function) (derivative outside function) ?

Not really sure what you're saying in the first line there. But I think what you want is this:

1. [(6x-4) - (6x+3] / (3x-2)^2 (this is what you have)

Multiply 1. by 3[(2x+1)/(3x-2)]^2 (you should see where this comes from).

I think this is right. At this point you can plug in a value of 1 for x and solve.

EDIT: It's been years since I did a differentiation this complex, so I could definitely be wrong.