Max volume problem, intro calc

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SUMMARY

The problem involves finding the dimensions of a rectangular box with the maximum volume that can be constructed from 100 square inches of cardboard, where the base length is twice the width. The volume formula used is V = lwh, which simplifies to V = 2n^2h given that the base length (l) is 2n and the width (w) is n. The surface area constraint is A = lw, which must equal 100 square inches, guiding the relationship between height and base dimensions.

PREREQUISITES
  • Understanding of volume and surface area formulas for rectangular prisms
  • Basic algebra for solving equations
  • Knowledge of optimization techniques in calculus
  • Familiarity with graphical representation of mathematical problems
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  • Study optimization techniques in calculus, focusing on finding maximum and minimum values
  • Learn how to apply constraints in optimization problems
  • Explore graphical methods for visualizing volume and surface area relationships
  • Practice similar problems involving dimensions and volume maximization
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Students studying calculus, particularly those focusing on optimization problems, as well as educators seeking to enhance their teaching methods in geometry and algebra.

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Homework Statement


What are the dimensions of the base of the rectangular box of the greatest volume that can be constructed from 100 sq inches of cardboard if the base is to be twice as long as it is wide? Assume the box has no top.


Homework Equations


V box = lwh
A = lw ?


The Attempt at a Solution


I did all the other similar problems assigned but am not sure where to begin here.

Drew a diagram. Base length= 2n width=n
volume=lwh
volume=2n*n*h
volume=2n^2*h

but now what. . .am I all wrong?

Sorry if this is way easy, it's just been awhile since I did these.
 
Physics news on Phys.org
Since all your material is going into making the box, the area of your material is the surface area of your box. Note that it doesn't have a top.
 

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