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

  1. #1
    Ife
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    [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?
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    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Ife View Post
    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.
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    Ife
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    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.
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    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Ife View Post
    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.
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    Ife
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    cool. thanx. i'm learning as i go along, still new to the input part of the forum.
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    Ife
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    Quote Originally Posted by VonNemo19 View Post
    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?
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    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Ife View Post
    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?
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    Ife
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    Quote Originally Posted by VonNemo19 View Post
    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.
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