Square-based pyramids surface area

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Discussion Overview

The discussion centers on determining which square-based pyramid with a fixed volume of V = 1 has the smallest surface area. The focus is on mathematical reasoning and calculus techniques to derive the surface area formula and analyze its minimum.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant proposes a method involving calculus to find the minimum surface area of square-based pyramids with a given volume.
  • Another participant questions a specific step in the derivation of the surface area formula, suggesting a potential error in the notation.
  • A subsequent reply acknowledges the oversight in notation but asserts that the final result remains valid.
  • Further discussion includes clarification on the mathematical operations involved, with participants confirming and correcting each other's interpretations of the formula.

Areas of Agreement / Disagreement

Participants engage in a debate regarding the correctness of specific mathematical steps, with some expressing uncertainty about the notation used in the derivation. No consensus is reached on the interpretation of the formula adjustments.

Contextual Notes

Limitations include potential misunderstandings of mathematical notation and the need for careful verification of each step in the calculus process. The discussion does not resolve these uncertainties.

arcnets
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Among all square-based pyramids which have volume V = 1, which one has the smallest surface area?
 
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This is a messy elementary calculus problem. Define:
a= length of base side
h= height of pyramid
t= altitude of triangle face
from this t2=h2+(a/2)2

Then

V (volume) = a2h/3
S (surface area) =a2+2at

Set V=1, use the above formula for t, and let x=a2, we get:

S=x+(36/x+x2)1/2

S'=1+(x-18/x2)/(36/x+x2)1/2

Set S'=0, we get x3=9/2

To verify that this is a minimum (not maximum or horizontal inflection), observe that:
x near 0, S approx 6/x1/2,
x gets large, S approx 2x.

Finally:
S=(9/2)1/3+((9/2)2/3+36(2/9)1/3)1/2

I suggest you work this through to understand how it goes.
 
Wow, mathman. Thanks.
I don't seem to understand this step:

S=x+(36/x+x2)1/2
S'=1+(x-18/x2)/(36/x+x2)1/2

Shouldn't it read

S'=1+(x-18/x2)/(36/x+x2)-1/2

...or something?
 
You're right. I had the minus when I was doing it (pencil and paper), but I neglected to type it in. However, the final result still stands.
 
Er, isn't it right as is?

dividing by z1/2 is the same as multiplying by z-1/2
 
Oops! Yes. I overlooked the /.
 
Hurkl got it right. When I read arcnets comment, I also forgot I had put in the /.
 

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