I have a problem!

I have attached it to the thread, just open the attachment please.

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- May 11th 2011, 02:43 AMemmaowenSmallest area and Smallest Length of String around Box.
I have a problem!

I have attached it to the thread, just open the attachment please. - May 11th 2011, 03:59 AMProve It
For starters, I think you'll find that the length of the string is actually 6x + 2y...

- May 11th 2011, 05:00 AMemmaowen
thankyou for that.

are you able to help any further?

Is my method ok besides my mistake?

How can i use it for the question i am stuck on, please? - May 11th 2011, 11:33 PMemmaowen
Please can anyone give me a few pointers??

- May 12th 2011, 03:45 AMemakarov
I think that in the problem you solved, the rope length is 4x + 2y, not 4x + 4y. Correspondingly, L = 4x + 2V / x^2, not 4x + V / x^2. Otherwise, I agree.

In the first problem, the variables m, w, and h seem independent, so the rope length has to be expressed through all three. Since V = 2mwh, you can express one of m, w, h through V and the other two and so express L through V and two of m, w, h. In particular, I don't see how you can find L in terms of V and w only. Also, minimizing the rope length requires some other constraint (e.g., keeping the volume constant); otherwise, you can get L = 0 when m = w= h = 0 . - May 12th 2011, 03:55 AMemmaowen
ahh sorry i have forgotten to say that volume = 3m^3 and w = 3m

so would you do something like this

L = 4w + 4m + 6H

For V = 2mwH make h the subject H = V/2mwH ??

Sub into equation

L = 4w + 4m + 3V/2mw - May 12th 2011, 04:14 AMemakarov
First, I get

L = 2(2h + 4m + 2w + 2h) = 8h + 8m + 4w. (*)

Here 2h corresponds to the front side, 4m + 2w to the top side, and 2h to the left side.

Second, V = 2mwh; substituting w = 3m, V = 6m^2h. Since also V = 3m^3, we get 2h = m. So, the side lengths are h, m = 2h and w = 3m = 6h. Also, V = 2mwh = 2 * 2h * 6h * h = 24h^3. So, everything can be expressed through h.

If you still need to find L in terms of w, then from (*) I get L = 8h + 8m + 4w = (8/6)w + (8/3)w + 4w. - May 12th 2011, 04:37 AMemmaowen
ahh oh i see gosh i was way off track... Just to clarify i don't fully understand how you did that last part when finding L in terms of w and volume?

Also, the question asks to find L in terms of w and Volume. I need V don't i? - May 12th 2011, 04:45 AMemakarov
Since w = 6h and w = 3m, we get h = w/6 and m = w/3. So, L = 8h + 8m + 4w = 8(w/6) + 8(w/3) + 4w = (4/3 +8/3 + 4)w = 8w.

I suggest you verify all calculations by doing them yourself without looking here because I make quite a few errors. - May 12th 2011, 04:53 AMemmaowen
Sorry to bother you further....

THe question asks to find the Length in terms of**Width and Volume**

i don't think this applies to you above solution or am i mistaken?

(oh and yes i do verify my all the calculations i have been nutting this question out on paper myself guided by your ideas in your threads) - May 12th 2011, 05:14 AMemakarovQuote:

THe question asks to find the Length in terms of Width and Volume

You gave two additional equations: volume = 3m^3 and w = 3m, which allowed reducing the number of independent variables. Initially, there are 5 variables: m, w, h, V, L and 2 equations: V = 2mwh and L = 8h + 8m + 4w. The number of independent variables is the total number of variables minus the number of equations (that can't be derived from each other), so initially there are 3 independent variables. When you added V = 3m^3 and w = 3m, the number of independent variables was reduced to 1. - May 12th 2011, 05:23 AMemmaowen
ok so if i was to draw a graph of v vs w how would that work?

- May 12th 2011, 05:28 AMemakarov
V = 2mwh = 2 * (w/3) * w * (w/6) = w^3 / 9. Here is the graph in WolframAlpha.

- May 12th 2011, 05:32 AMemmaowen
Ah i see... Thankyou so much for your time. This will help me to practice other questions similar

Thanks again