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anonymous

  • 5 years ago

A rectangular 48" by 36" piece of cardboard is used to create an open box by cutting a square from each corner and folding up the sides. If one of the dimensions of the box is to be 20 inches what is the largest possible volume of the box?

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  1. anonymous
    • 5 years ago
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    Are you in calculus?

  2. anonymous
    • 5 years ago
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    no Algebra honors

  3. anonymous
    • 5 years ago
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    Hrm... This is a tough one, since drawing it out would be very beneficial as well. https://docs.google.com/drawings/edit?id=1crhrg5jI196FHJNvCH64xY02Ze_3EeObtkyAhBGV9zY&hl=en&authkey=CPGXwM4O Okay, so we can setup the equation now as such: (48-2x)+(36-2x)=V V=2x+2y Using both equations we should be able to solve for the max possible volume of the box.

  4. anonymous
    • 5 years ago
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    I got y= 42-3x?

  5. anonymous
    • 5 years ago
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    yup

  6. anonymous
    • 5 years ago
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    so what do I do with that ?

  7. anonymous
    • 5 years ago
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    Well we know that x is one length and y is another. Since we solved for y, we can substitute x=20, and solve for y. Does it make sense why?

  8. anonymous
    • 5 years ago
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    yess ok so now I got 18 for y

  9. anonymous
    • 5 years ago
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    so that should be the max for the second dimension when creating an open box like this.

  10. anonymous
    • 5 years ago
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    Sweet thanks :)

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