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Optimization - Volume of a Box

  • Thread starter roman15
  • Start date
  • #1
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Homework Statement



Ok I know this question is really easy but for some reason I got it wrong.

You are given a piece of sheet metal that is twice as long as it is wide and has an area of 800m^2. Find the dimensions of the rectangular box that would contain a maximum volume if it were constructed from this piece of metal by cutting squares of equal area at all four corners and folding up the sides. The Box will not have a lid.


Homework Equations





The Attempt at a Solution


So basically I drew the piece of metal, one side was 2m the other was 1m, then I subtracted the corners, so I had 2-2x=1-x and 1-2x and the height being x
so V=x(1-x)(1-2x)
=2x^3-3x^2+x
then V'=6x^2-6x+1
but that didnt give me the right answer...Im not sure, but does the area have any significance to the problem?
 

Answers and Replies

  • #2
23
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Yes. All of your logic for the calculus problem was right, except for the fact that one side does not equal 2m and the other does not equal 1 m. You have to use algebra to figure you the length and width of the rectangular box.

width = w
length = l
l = 2w
800 = 2w^2
400 = w^2
w = 20
l = 40

So the equation above should be
V = x(40-2x)(20-2x)
 

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