Maximizing the volume of a square Pyramid?

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Homework Help Overview

The problem involves maximizing the volume of a square-based pyramid constructed from an 8cm by 8cm piece of paper. The relevant equations for volume and surface area are provided, indicating a relationship between the base dimensions and height of the pyramid.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the base size and volume, noting that larger bases may lead to larger volumes, while larger altitudes may reduce volume. There is mention of using calculus to find the maximum volume and the need to establish a formula that relates volume to a chosen parameter.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of how to approach the problem. Some have suggested that the surface area constraint must be considered, while others are focused on the relationship between base size and volume. There is no explicit consensus yet on the best method to maximize the volume.

Contextual Notes

Participants note that the surface area of the pyramid cannot exceed 64 cm², which may influence the dimensions of the pyramid being constructed. There is also a consideration of how much material is cut from the square base.

agv567
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Homework Statement



I am given an 8cm by 8cm piece of paper. I have to cut out a square-based pyramid out of that that gives me the greatest volume.

Homework Equations



I know that the volume is V = 1/3 * b^2 * h
b = base
h = height


I know that the surface area is A = b^2 + 2bh


The Attempt at a Solution



I slightly remember optimization. I only remember you need to combine the two but that's it?
 
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agv567 said:

Homework Statement



I am given an 8cm by 8cm piece of paper. I have to cut out a square-based pyramid out of that that gives me the greatest volume.

Homework Equations



I know that the volume is V = 1/3 * b^2 * h
b = base
h = height

I know that the surface area is A = b^2 + 2bh

The Attempt at a Solution



I slightly remember optimization. I only remember you need to combine the two but that's it?
The surface area isn't involved in this solution. Of course I realize that the surface area of the pyramid can't be greater than 64 cm2.

Connect the four midpoints of the edges of the sheet of paper. Let the resulting square be the base of a pyramid. Fold the corner triangles upward & inward until the corners of the original 8×8 piece of paper meet at the apex of a pyramid. That square pyramid is one with maximum base area. You would hardly call it a pyramid at all, because its altitude is zero.

Other square pyramids will require that you use a smaller base, but will result in larger altitude. Such pyramids will have surface areas less than 64 cm2.
 
yeah i experimented and found out that the smaller you cut from the square, the larger the volume(aka larger base = larger volume). Larger altitudes give you smaller volumes.

How would I find out the maximum volume using calculus work though?

l
 
agv567 said:
yeah i experimented and found out that the smaller you cut from the square, the larger the volume(aka larger base = larger volume). Larger altitudes give you smaller volumes.

How would I find out the maximum volume using calculus work though?
Come up with a formula relating the volume to some parameter related to how much is cut out, or related to the length of an edge of the square base, or related to the area of the base, or related to ...

So the volume will be a function of whatever parameter you choose.

Then use calculus to find the maximum value of that function.
 

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