Maximum dimensions-very interesting

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SUMMARY

The discussion focuses on optimizing the volume of an open-top box created from a square cardboard sheet with an area of 1600 cm². The cardboard has a side length of 40 cm, and the corners are cut out to form a square base. The maximum volume is achieved when the side length of the corner square is set to 20/3 cm, resulting in a base of 80/3 cm and a height of 20/3 cm. The critical points for volume are determined by setting the derivative of the volume function, V = (40-2a)² * a, to zero.

PREREQUISITES
  • Understanding of basic calculus, specifically derivatives
  • Familiarity with volume calculations for three-dimensional shapes
  • Knowledge of optimization techniques in mathematics
  • Ability to manipulate algebraic expressions
NEXT STEPS
  • Study calculus applications in optimization problems
  • Learn about the geometric properties of three-dimensional shapes
  • Explore the use of derivatives in finding maxima and minima
  • Investigate real-world applications of volume optimization in packaging design
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Students in mathematics or engineering fields, educators teaching optimization techniques, and anyone interested in practical applications of calculus in design and manufacturing.

joejo
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maximum dimensions-very interesting!

hey guys...i have another question...i posted my answer right below it...does it look right guys?! I hope so! Thanks in advance

You are to design a container box by cutting out the four corners of a square cardboard sheet that is 1600cm^2 in area. The box must have a square base and an open top. Determine the dimensions of the box that give maximum value.

The side of the square cardboard is 40 cm. Let side of corner square is a cm. The volume of the box V = (40-2a)2.a cm3 , For max volume we have dV/da = 0 implies (20-3a)(20-a) = 0, i.e., a = 20, a = 20/3. But when a = 20, V = 0. So to maximize the volume a = 20/3. So the base of is a square of side 80/3 cm and height = 20/3 cm.

Sorry not good with latex...
 
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Yes, that looks good. It's intuitively clear that a=20/3 should be a maximum (there must be 'some' maximum between a=0 and a=20), but for completeness you might want to show it is a maximum indeed (depends on your teacher).
 
thanks for your help!
 

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