Optimization - Volume of a Box

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SUMMARY

The discussion focuses on optimizing the volume of an open-top rectangular box constructed from a piece of sheet metal that is twice as long as it is wide, with a total area of 800 m². The correct dimensions are derived using algebra, where the width (w) is determined to be 20 m and the length (l) is 40 m. The volume formula is established as V = x(40 - 2x)(20 - 2x), where x represents the height of the box. The initial miscalculations were clarified, emphasizing the importance of correctly determining the dimensions based on the area provided.

PREREQUISITES
  • Understanding of algebraic equations and variables
  • Knowledge of calculus, specifically differentiation for optimization
  • Familiarity with geometric concepts related to volume
  • Basic understanding of area calculations for rectangles
NEXT STEPS
  • Study optimization techniques in calculus, focusing on finding maximum and minimum values
  • Learn about the application of derivatives in real-world problems
  • Explore geometric properties of rectangular prisms and their volume calculations
  • Investigate the relationship between area and dimensions in geometric shapes
USEFUL FOR

Students studying calculus, particularly those focusing on optimization problems, as well as educators teaching geometric volume concepts and algebraic manipulation.

roman15
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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?
 
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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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