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Math Help - Least amount of materials problem

  1. #1
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    Least amount of materials problem

    I've been given a problem and would like to see if I have thought correctly when solving it. A box without lid with a volume of 4dm3 should have as small area as possible for the 5 limiting sides. the base should be square.

    My solution: The base area is b2 and the area for the 4 sides are b*h so total side area is 4*b*h. For the volume 4 we get that b2*h = 4 so h = 4/b2.
    The total area is A = b2+4*b*(4/b2) -> A = b2 + 16/b.

    Derivating and skissing the curve I found b=2 to be a minimum so with b=2 would therefor result in the least amount of materials used. Is this the correct steps for solving this type of problem?
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  2. #2
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    Re: Least amount of materials problem

    Quote Originally Posted by dipsy34 View Post
    I've been given a problem and would like to see if I have thought correctly when solving it. A box without lid with a volume of 4dm3 should have as small area as possible for the 5 limiting sides. the base should be square.

    My solution: The base area is b2 and the area for the 4 sides are b*h so total side area is 4*b*h. For the volume 4 we get that b2*h = 4 so h = 4/b2.
    The total area is A = b2+4*b*(4/b2) -> A = b2 + 16/b.

    Derivating and skissing the curve I found b=2 to be a minimum so with b=2 would therefor result in the least amount of materials used. Is this the correct steps for solving this type of problem?
    For a minimum require dA/db =0 Solve this and you will indeed get b=2
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  3. #3
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    Re: Least amount of materials problem

    Quote Originally Posted by dipsy34 View Post
    I've been given a problem and would like to see if I have thought correctly when solving it. A box without lid with a volume of 4dm3 should have as small area as possible for the 5 limiting sides. the base should be square.

    My solution: The base area is b2 and the area for the 4 sides are b*h so total side area is 4*b*h. For the volume 4 we get that b2*h = 4 so h = 4/b2.
    The total area is A = b2+4*b*(4/b2) -> A = b2 + 16/b.

    Derivating and skissing the curve I found b=2 to be a minimum so with b=2 would therefor result in the least amount of materials used. Is this the correct steps for solving this type of problem?
    I have no idea what "skissing the curve" could mean! b= 2 is correct- but you would still need to determine h from 4/b^2. But we can't tell whether you used the "correct steps" until you tell us what steps you took.
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