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Math Help - Applied Minimum

  1. #1
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    Applied Minimum

    A container manufacturer is designing a retagular box, open at the top with a square base, that is to have a volume of 32 ft^3. If the box is to require the least amount of materials, what would be its dimensions?

    I know that the answer to this equation is 4X4X2 but I'm not sure what equation to use to get this answer. I started by doing...

    32= L*W*H
    32= 2y*X
    y=32-X/2

    Am I on the right track or is there another equation I am suppost to use. If anyone know the answer could you please let me know. Thank you so much.
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  2. #2
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    Quote Originally Posted by johnkw87 View Post
    A container manufacturer is designing a retagular box, open at the top with a square base, that is to have a volume of 32 ft^3. If the box is to require the least amount of materials, what would be its dimensions?

    I know that the answer to this equation is 4X4X2 but I'm not sure what equation to use to get this answer. I started by doing...

    32= L*W*H
    32= 2y*X ... no
    y=32-X/2
    square base with side length x, height of box h ...

    32 = x^2h

    surface area ...

    A = x^2 + 4xh

    using the volume formula, solve for h in terms of x and substutute the result into the surface area equation ... find dA/dx and minimize
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