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anonymous

  • 5 years ago

A container with a square base, vertical sides and an open top is to be made from 1200ft squared of material. find the dimensions of the container with the greatest volume.

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  1. anonymous
    • 5 years ago
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    how do i begin

  2. amistre64
    • 5 years ago
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    Do you know if all 1200 ft^2 of material is to be used? or is it one of those"cut out the corners" kind of problems?

  3. amistre64
    • 5 years ago
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    the container, if all material is to be used would have a surface area of: 4(side area) + (base area) = 1200

  4. anonymous
    • 5 years ago
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    you just need the sum of the areas of the surfaces to add up to 1200sqft. so (area of base)+4×(area of one side) = 1200 sqft

  5. amistre64
    • 5 years ago
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    4xy + x^2 = 1200 , does that sound right?

  6. amistre64
    • 5 years ago
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    Volume = x^2 y

  7. amistre64
    • 5 years ago
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    solve for one variable in terms of the other, plug it into the volume equation, and derive :)

  8. amistre64
    • 5 years ago
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    y = (1200 -x^2)/4x

  9. amistre64
    • 5 years ago
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    and dont let the "hero" title fool ya.... I am after all an idiot in disguise :)

  10. amistre64
    • 5 years ago
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    V = (x^2/4x)(1200-x^2)

  11. amistre64
    • 5 years ago
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    V = (1/4) (1200-x^3) if I did it right...

  12. amistre64
    • 5 years ago
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    *1200x - x^3

  13. amistre64
    • 5 years ago
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    dV = (1/4)(1200 - 3x^2) maybe make that equal to 0 to get the critical numbers

  14. anonymous
    • 5 years ago
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    Oh, I misread the problem.. yes, put y in terms of x, y = (1200 -x^2)/4x, then substitute into volume V = x^2 y = x^2 (1200 -x^2)/4x = x(1200-x^2)/4 = x(300-(x^2)/4), then find the maximum volume

  15. amistre64
    • 5 years ago
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    1200 = 3x^2

  16. amistre64
    • 5 years ago
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    x = sqrt400)

  17. anonymous
    • 5 years ago
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    If you've taken calculus you can do that my finding the zeros of the derivative, otherwise I think you'll have to test sample values and just find the x that yields the greatest V

  18. amistre64
    • 5 years ago
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    x=20 i think plug that back in to your "y" equation to ge the value for y

  19. amistre64
    • 5 years ago
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    y = (1200 -20^2)/4(20) y = (1200 - 400)/ 80 y = 800/80 = 80/8 = 10/1 y = 10 and x = 20 if i did it all correctly

  20. amistre64
    • 5 years ago
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    determine if 20x20x10 is greater than 10x10x20 and youve got your answer

  21. amistre64
    • 5 years ago
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    or.... ignore that last comment...its probably my stupidity talking :)

  22. amistre64
    • 5 years ago
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    20x20 for the base, and 10 high fits all the requirements

  23. amistre64
    • 5 years ago
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    but, do you see how we got it?

  24. anonymous
    • 5 years ago
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    yeah

  25. anonymous
    • 5 years ago
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    square roots :)

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