# Thread: Smallest area and Smallest Length of String around Box.

1. ## Smallest area and Smallest Length of String around Box.

I have a problem!

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

2. For starters, I think you'll find that the length of the string is actually 6x + 2y...

3. 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?

4. Please can anyone give me a few pointers??

5. 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 .

6. 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

7. 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.

8. 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?

9. 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.

10. 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)

11. THe question asks to find the Length in terms of Width and Volume
Well, L = 8w satisfies this requirement; it just does not use 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.

12. ok so if i was to draw a graph of v vs w how would that work?

13. V = 2mwh = 2 * (w/3) * w * (w/6) = w^3 / 9. Here is the graph in WolframAlpha.

14. Ah i see... Thankyou so much for your time. This will help me to practice other questions similar
Thanks again