Square-based pyramids surface area

1. May 28, 2003

arcnets

Among all square-based pyramids which have volume V = 1, which one has the smallest surface area?

2. May 31, 2003

mathman

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.

3. Jun 1, 2003

arcnets

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

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

...or something?

4. Jun 1, 2003

mathman

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.

5. Jun 1, 2003

Hurkyl

Staff Emeritus
Er, isn't it right as is?

dividing by z1/2 is the same as multiplying by z-1/2

6. Jun 2, 2003

arcnets

Oops! Yes. I overlooked the /.

7. Jun 2, 2003

mathman

Hurkl got it right. When I read arcnets comment, I also forgot I had put in the /.