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zeesbrat3

  • one year ago

A box is to be constructed from a sheet of cardboard that is 20 cm by 60 cm, by cutting out squares of length x by x from each corner and bending up the sides. What is the maximum volume this box could have?

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  1. zeesbrat3
    • one year ago
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    @ganeshie8

  2. zeesbrat3
    • one year ago
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    @jim_thompson5910

  3. zeesbrat3
    • one year ago
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    @freckles

  4. jim_thompson5910
    • one year ago
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    |dw:1436828452402:dw|

  5. jim_thompson5910
    • one year ago
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    if we make the cuts and fold up the box, we get something like this |dw:1436828568511:dw|

  6. zeesbrat3
    • one year ago
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    Okay. What next? I know Volume would be lwh --- so x(20-2x)(60-2x)?

  7. jim_thompson5910
    • one year ago
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    correct

  8. zeesbrat3
    • one year ago
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    So, then what?

  9. jim_thompson5910
    • one year ago
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    ok so if you graph x(20-2x)(60-2x) with something like desmos, you get this (see attached)

  10. zeesbrat3
    • one year ago
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    Okay?

  11. jim_thompson5910
    • one year ago
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  12. zeesbrat3
    • one year ago
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    So 2525 is the max volume?

  13. jim_thompson5910
    • one year ago
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    desmos is saying that the max point is (4.514, 2525) so x = 4.514 and the max volume is 2525 yes

  14. zeesbrat3
    • one year ago
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    Thank you!

  15. jim_thompson5910
    • one year ago
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    no problem

  16. zeesbrat3
    • one year ago
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    So what if you wanted to work it out by hand? Just wondering..

  17. jim_thompson5910
    • one year ago
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    to do so by hand, you would need to know calculus (specifically, derivatives)

  18. zeesbrat3
    • one year ago
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    I do haha. I did it originally using derivatives and my teacher yelled at me and marked it wrong

  19. jim_thompson5910
    • one year ago
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    such a strange thing to be yelled at for lol

  20. jim_thompson5910
    • one year ago
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    I'm guessing you're in precalculus and the teacher wanted you to use technology like a graphing calculator

  21. zeesbrat3
    • one year ago
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    I'm actually in calculus and my sister taught me derivatives

  22. jim_thompson5910
    • one year ago
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    ok let's use derivatives then

  23. jim_thompson5910
    • one year ago
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    first expand out x(20-2x)(60-2x) to get x(20-2x)(60-2x) = x(1200 - 160x + 4x^2) = 4x^3 - 160x^2 + 1200x

  24. jim_thompson5910
    • one year ago
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    then you apply a derivative to 4x^3 - 160x^2 + 1200x to get y = 4x^3 - 160x^2 + 1200x dy/dx = 3*4x^(3-1) - 2*160x^(2-1) + 1*1200x^(1-1) dy/dx = 12x^2 - 320x + 1200

  25. zeesbrat3
    • one year ago
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    Ya, I got that far. Then quadratic?

  26. jim_thompson5910
    • one year ago
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    Now you set the derivative 12x^2 - 320x + 1200 equal to zero and solve for x 12x^2 - 320x + 1200 = 0 x = 22.15250437 or x = 4.514162294 I skipped a bunch of steps since I'm sure you know the quadratic formula very well at this point

  27. zeesbrat3
    • one year ago
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    Then plug them back in for the y

  28. jim_thompson5910
    • one year ago
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    since the width is 20-2x, this means that 20-2x must be positive 20 - 2x > 0 20 > 2x 2x < 20 x < 10 so x is both positive, x > 0, and less than 10. Put together, 0 < x < 10 is the domain restriction here. This is visually confirmed in that desmos graph above

  29. jim_thompson5910
    • one year ago
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    because 0 < x < 10, we can ignore x = 22.15250437

  30. zeesbrat3
    • one year ago
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    I find it extremely frightening that you just completed the problem so quickly going the long way

  31. jim_thompson5910
    • one year ago
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    I've had lots of practice

  32. zeesbrat3
    • one year ago
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    Maybe one day I will be able to do that

  33. jim_thompson5910
    • one year ago
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    you will with enough practice

  34. zeesbrat3
    • one year ago
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    Again, thank you for your help

  35. jim_thompson5910
    • one year ago
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    no problem

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