Among all square-based pyramids which have volume V = 1, which one has the smallest surface area?
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
V (volume) = a2h/3
S (surface area) =a2+2at
Set V=1, use the above formula for t, and let x=a2, we get:
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.
I suggest you work this through to understand how it goes.
Wow, mathman. Thanks.
I don't seem to understand this step:
Shouldn't it read
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 /.
Separate names with a comma.