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anonymous

  • 5 years ago

Let's construct a box whose base length is 3 times its base width and will have no lid. We only have 64 m2 of material to use. Determine the dimensions of the box that will give the largest volume. ___width? ___height? ___length? What is the volume of the box? ANY HELP ON HOW TO START THIS PLEASE??

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  1. amistre64
    • 5 years ago
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    we know that the surface area has to equal 64m^2 right?

  2. amistre64
    • 5 years ago
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    our dimensions can be l,w,h were l = 3w

  3. amistre64
    • 5 years ago
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    surface area = 3w(w) + 2(hw) + 2(3wh) = 64

  4. amistre64
    • 5 years ago
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    volume of the box = 3w(w)h

  5. amistre64
    • 5 years ago
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    64 = 3w^2 +2wh +6wh 64 = 3w^2 +8wh solve for "h" 64 - 3w^2 = 8wh 64-3w^2 -------- = h 8w

  6. amistre64
    • 5 years ago
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    use this "value" of h in the formula for the volume: V = 3w^2 [(64-3w^2)/8w]

  7. amistre64
    • 5 years ago
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    V = (192w^2 - 9w^4)/8w ; now simplify

  8. amistre64
    • 5 years ago
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    V = (192x -9w^3)/8 find the derivative of this Volume function to determine critical numbers

  9. amistre64
    • 5 years ago
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    8(192-27w^2) ------------- 8(8) 192-27w^2 ---------- solve for zero 8 192 - 27w^2 = 0 gonna be tricky for me without pencil and paper :)

  10. amistre64
    • 5 years ago
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    got it.. 3(64-9w^2) = 0 64-9w^2 = 0 (8+3w)(8-3w) = 0 w = -3/8 or w = 3/8 since a negative width is meaningless; lets use w = 3/8

  11. amistre64
    • 5 years ago
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    recall that l = 3w: l = 3(3/8) = 9/8

  12. amistre64
    • 5 years ago
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    64-3w^2 -------- = h 8w 64 - 3(3/8)^2 ------------ = h 8(3/8) 64 - (27/64) ------------ = h 3

  13. amistre64
    • 5 years ago
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    4069/64 -------- = h 3 4069/192 = h might need to reduce that....

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