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

1. ## [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?

2. 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 $width=height$. If this is the case, wwe develop the two equations

$V=lwh=2lw$

And

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

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

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

Differentiating and maximizing

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

You can do the rest.

3. 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.

4. 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.

5. cool. thanx. i'm learning as i go along, still new to the input part of the forum.

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

$V=lwh=2lw$

And

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

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

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

Differentiating and maximizing

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

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?

7. 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?

8. 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!

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