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anonymous

  • 5 years ago

Caculus help please! A box with an open top is to be constructed from a rectangular piece of cardboard with dimensions b=8in by a=10in by cutting out equal squares of side x at each corner and then folding up the sides how much do we cut from each corner in order to maximize the volume?

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  1. radar
    • 5 years ago
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    Let x = the dimension of the corner cut. This is also the height.

  2. radar
    • 5 years ago
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    let lenght =10-2x width = 8-2x Volume =l x w x h V = x (8-2x)(10-2x) Complete and differentiae set to 0 and solve. got to run good luck.

  3. myininaya
    • 5 years ago
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    ok so V'(x)=1*(8-2x)(10-2x)+x(-2)(10-2x)+x(8-2x)(-2)

  4. myininaya
    • 5 years ago
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    we can clean that up so finding x from V'=0 is easier

  5. myininaya
    • 5 years ago
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    V'(x)=80-16x-20x+4x^2-20x+4x^2-16x+4x^2

  6. myininaya
    • 5 years ago
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    we have like terms

  7. myininaya
    • 5 years ago
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    V'(x)=8x^2-72x+80 unless i made a mistake

  8. myininaya
    • 5 years ago
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    V'(x)=8(x^2-9x+10)

  9. myininaya
    • 5 years ago
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    i did make a mistake V'=12x^2-72x+80

  10. myininaya
    • 5 years ago
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    now set V'=0 to find critical numbers let me know if you have anymore trouble

  11. anonymous
    • 5 years ago
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    this helps a lot but i can't factor 12x^2-72x+80=0. and to find x: the cut out corners are x by x

  12. radar
    • 5 years ago
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    Divide all terms by 4 getting \[3x ^{2}-18x+20=0\]

  13. radar
    • 5 years ago
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    a=3, b=-18, and c\[(18\pm \sqrt{84})/6\]

  14. radar
    • 5 years ago
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    \[(18\pm2\sqrt{21})/6\] Results in x approx 1.4724 in.

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