Square-based pyramids surface area

[arcnets]

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

 

[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/x^1/2,
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.

 

[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

Shouldn't it read

S'=1+(x-18/x²)/(36/x+x²)^-1/2

...or something?

 

[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.

 

[Hurkyl]

Er, isn't it right as is?

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

 

[arcnets]

Oops! Yes. I overlooked the /.

 

[mathman]

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