anonymous
  • anonymous
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?
Mathematics
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anonymous
  • anonymous
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?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
Are you in calculus?
anonymous
  • anonymous
no Algebra honors
anonymous
  • anonymous
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.

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anonymous
  • anonymous
I got y= 42-3x?
anonymous
  • anonymous
yup
anonymous
  • anonymous
so what do I do with that ?
anonymous
  • anonymous
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?
anonymous
  • anonymous
yess ok so now I got 18 for y
anonymous
  • anonymous
so that should be the max for the second dimension when creating an open box like this.
anonymous
  • anonymous
Sweet thanks :)

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