## anonymous one year ago optimization problem. Maximum volume of a box? I don't know how to set this up. We're using a paper with dimensions 8(1/2) inches by 11 inches to form a box WITH a lid with maximum volume. Obviously I'm using the formula v=length x width x height but I don't know how to get v(x) :/

1. dan815

|dw:1433466000616:dw|

2. dan815

also ask yourself are you able to cut the paper in any way u want or do u have to fold it

3. dan815

is it possible to just wrap this paper around some cube?

4. dan815

any thoughts so far?

5. dan815

|dw:1433466271247:dw|

6. anonymous

ok so I'm able to cut identical squares out of the corners and in the middle so I cold then fold it up into the box.

7. ybarrap

You might want to try using the method of Lagrange: http://en.wikipedia.org/wiki/Lagrange_multiplier $$\text{Maximize: }l\times w\times h\\ \text{Subject to the constraint: }2\times(hw+lh+wh)=8.5\times 11\\ \text{Hence,}\\ \Lambda(x,y,z,\lambda)=lwh+\lambda\left (hw+lh+wl-\cfrac{8.5\times 11}{2}\right )$$ Now find the partials: $$\Lambda_h=wl+\lambda(w+l)=0\\ \Lambda_w=hl+\lambda(h+l)=0\\ \Lambda_l=hw+\lambda(h+w)=0\\ \Lambda_{\lambda}=hw+lh+wl-\cfrac{8.5\times 11}{2}=0\\$$ Using first equation, we find for $$\lambda$$ $$\lambda=-\cfrac{wl}{w+l}$$ Plugging this into the second and third equation we get that $$w=l$$ Using this result, we get that $$\lambda =-\cfrac{w}{2}$$ Using these results into the last two equations we find the final value for $$h$$. I'll let you do that. Once you have this result, plug back into your constraint to find the actual values. Does this make sense?

8. anonymous

I had to reread it a few times but I think I got it. Thank you :)

9. ybarrap

You're welcome