Square-based pyramids surface area

  • Thread starter arcnets
  • Start date
512
0
Among all square-based pyramids which have volume V = 1, which one has the smallest surface area?
 

mathman

Science Advisor
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389
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.
 
512
0
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?
 

mathman

Science Advisor
7,688
389
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.
 

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
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Er, isn't it right as is?

dividing by z1/2 is the same as multiplying by z-1/2
 
512
0
Oops! Yes. I overlooked the /.
 

mathman

Science Advisor
7,688
389
Hurkl got it right. When I read arcnets comment, I also forgot I had put in the /.
 

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