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anonymous

  • one year ago

from a thin piece of cardboard 10 in. by 10in, square corners are cut out so that the sides can be folded up to make a box. what dimensions will yield a box of maximum volume? what is the maximum volume?

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  1. misty1212
    • one year ago
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    |dw:1438826627031:dw|

  2. misty1212
    • one year ago
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    each side is \(10-2x\) and the height is \(x\) so the volume is \[V(x)=(10-2x)^2x\]

  3. misty1212
    • one year ago
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    multiply this mess out , get \[V(x)=4 x^3-40 x^2+100 x\]

  4. misty1212
    • one year ago
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    then take the derivative, set it equal to zero, solve for \(x\) to get your answer

  5. misty1212
    • one year ago
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    you good from there?

  6. anonymous
    • one year ago
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    ahh I see thank you so much

  7. anonymous
    • one year ago
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    you get two values of x find the second derivative and see at which point it is negative,that value of x gives maximum volume.

  8. misty1212
    • one year ago
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    make sure you pick a root that makes sense

  9. misty1212
    • one year ago
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    don't forget that the domain of your function is only \((0,5)\)

  10. misty1212
    • one year ago
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    i have an answer if you want me to check yours, let me know

  11. anonymous
    • one year ago
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    ok so I got 12x^2-80x-100 for the derivative so then I pulled out a gcf 4x(3x-20)=0 and then x=20/3

  12. mathstudent55
    • one year ago
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    f(x) = 4x^3 - 40x^2 + 100x f'(x) = 12x^2 - 80x + 100 Check your derivative.

  13. anonymous
    • one year ago
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    so don't pull out a gcf? im confused

  14. anonymous
    • one year ago
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    so do I use quadratic equation to solve this problem is stupid

  15. mathstudent55
    • one year ago
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    You wrote the derivative with a last term of -100. I'm telling you above that the last term of the derivative is + 100. Now solve 12x^2 - 80x + 100 = 0

  16. anonymous
    • one year ago
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    Refer to the attachment.

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