Volume of an Open-Top Box with 3-Sided Squares

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SUMMARY

The discussion centers on the mathematical problem of determining the volume of an open-top box formed from a rectangular piece of cardboard by cutting squares from the corners. The key equations derived are V = (x-2)(y-2) and 1 = 2(x+2)(y+2), which relate the dimensions of the cardboard to the volumes produced by cutting squares of different sizes. The analysis reveals that the relationship between the dimensions x and y is crucial, leading to the equations 0 = xy + 6x + 6y + 28 and 16x + 16y + 63 = 0, which are essential for further exploration of the problem.

PREREQUISITES
  • Understanding of algebraic manipulation and equations
  • Familiarity with volume calculations for geometric shapes
  • Knowledge of the properties of rectangular dimensions
  • Experience with factorization techniques in algebra
NEXT STEPS
  • Explore the implications of the equations 0 = xy + 6x + 6y + 28 and 16x + 16y + 63 = 0
  • Investigate the geometric properties of open-top boxes and their volume calculations
  • Learn about systems of equations and their applications in geometry
  • Study factorization methods and their relevance in solving polynomial equations
USEFUL FOR

This discussion is beneficial for students studying algebra, educators teaching geometric volume concepts, and anyone interested in mathematical problem-solving involving dimensions and volumes of shapes.

camilus
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Homework Statement



An open-top box can be formed from a rectangular piece of cardboard by cutting equal squares from the four corners and then folding up the four sections that stick out. For a particular-sized piece of cardboard, the same volume results whether squares of side one or squares of side two have been cut out. What is the resulting volume if squares of side three are cut out?
 
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More information is needed. Is there any relation between the sides of the rectangle? Any perimeters? any areas? Just the equality of two particular starter volumes gives this:

V = (x-2)(y-2) = 2(x-4)(y-4)
and then
steps of algebra,
0 = xy + 6x + 6y + 28, apparently not factorable, not useful alone;

Or since the second volume expression is factorable,
(x-2)(y-2) = 2(x+2)(x-2)(y+2)(y-2)
which yields
1 = 2(x+2)(y+2)

Interesting. That gives two different possible useable equations in x and y:
0=xy+6x+6y+28
AND
1=2xy+4x+4y+8
 
With continued steps, I obtain
16x + 16y + 63 =0
 

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