# [SOLVED] Dimensions of a box for shipping.. i think this has to do with Limits, not s

• Dec 5th 2009, 03:24 PM
Ife
[SOLVED] Dimensions of a box for shipping.. i think this has to do with Limits, not s
I have a few worded problems i have no clue how to go about solving... there are diagrams, but i don't think i can draw it online, but i think the question is a bit explanitory, so lets see how it goes.

A private company will accept a box for domestic shipment only if the sum of its length and girth (distance around) does not exceed 102 in. What dimensions will give a box with a square end the largest possible volume?
• Dec 5th 2009, 04:32 PM
VonNemo19
Quote:

Originally Posted by Ife
I have a few worded problems i have no clue how to go about solving... there are diagrams, but i don't think i can draw it online, but i think the question is a bit explanitory, so lets see how it goes.

A private company will accept a box for domestic shipment only if the sum of its length and girth (distance around) does not exceed 102 in. What dimensions will give a box with a square end the largest possible volume?

I think that what the problem is saying is that $\displaystyle width=height$. If this is the case, wwe develop the two equations

$\displaystyle V=lwh=2lw$

And

$\displaystyle 102in=l+w+h=l+w+w=l+2w$

Now, solving for $\displaystyle w$ in the second equation and then substituting into the first gives

$\displaystyle V(l)=2l\frac{102-l}{2}=l(102-l)=102l-l^2$

Differentiating and maximizing

$\displaystyle V'(l)=102-2l=0\Rightarrow{l}=51in$

You can do the rest.
• Dec 5th 2009, 05:15 PM
Ife
Ok i was thinking along the right track with the width=height part. I just wasnt sure where to take it from there. Thanks a lot! I hope the diagram helped, i tried hard to get it like on the question. I'll work the question from that direction you gave me.
• Dec 5th 2009, 05:35 PM
VonNemo19
Quote:

Originally Posted by Ife
Ok i was thinking along the right track with the width=height part. I just wasnt sure where to take it from there. Thanks a lot! I hope the diagram helped, i tried hard to get it like on the question. I'll work the question from that direction you gave me.

Hey Ife, I like you so I'm gonna give it to you straght. Try not to attach files that require downloading. This scares people away from helping you out.

Peace out Fresh. (Wink)
• Dec 5th 2009, 06:09 PM
Ife
cool. thanx. i'm learning as i go along, still new to the input part of the forum. (Doh)
• Dec 10th 2009, 07:27 PM
Ife
Quote:

Originally Posted by VonNemo19
I think that what the problem is saying is that $\displaystyle width=height$. If this is the case, wwe develop the two equations

$\displaystyle V=lwh=2lw$

And

$\displaystyle 102in=l+w+h=l+w+w=l+2w$

Now, solving for $\displaystyle w$ in the second equation and then substituting into the first gives

$\displaystyle V(l)=2l\frac{102-l}{2}=l(102-l)=102l-l^2$

Differentiating and maximizing

$\displaystyle V'(l)=102-2l=0\Rightarrow{l}=51in$

You can do the rest.

Thanks, but 1 thing: shouldn't the 102 = l+4w and not 2w?? since the girth is the distance all around, so that would be the sum of the 4 sides of the same length + the length of the box?
• Dec 10th 2009, 08:10 PM
VonNemo19
Quote:

Originally Posted by Ife
Thanks, but 1 thing: shouldn't the 102 = l+4w and not 2w?? since the girth is the distance all around, so that would be the sum of the 4 sides of the same length + the length of the box?

You know what? I think that you might be right. Good eye Ife. I'm sorry if I caused any confusion. With this information, do you know how to proceed?
• Dec 10th 2009, 08:12 PM
Ife
Quote:

Originally Posted by VonNemo19
You know what? I think that you might be right. Good eye Ife. I'm sorry if I caused any confusion. With this information, do you know how to proceed?

Thanks! :D

And nah, no confusion caused. I am still eternally grateful for your assistance.