This is a messy elementary calculus problem. Define:
a= length of base side
h= height of pyramid
t= altitude of triangle face
from this t^{2}=h^{2}+(a/2)^{2}
Then
V (volume) = a^{2}h/3
S (surface area) =a^{2}+2at
Set V=1, use the above formula for t, and let x=a^{2}, we get:
S=x+(36/x+x^{2})^{1/2}
S'=1+(x-18/x^{2})/(36/x+x^{2})^{1/2}
Set S'=0, we get x^{3}=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.
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