math

Two right circular cone, one upside down in the other. The two bases are parallel. The vertex of the smaller cone lies at the center of the larger cone’s base. The larger cone’s height and base radius are 12 and 16 ft, respectively. What are the smaller cone’s height (h) and base radius (r), both are in ft, that maximize the smaller cone’s volume?

R = radius of bigger cone = 6 ft
r = radius of smaller cone…?
H = height of bigger cone = 12 ft
h = height of smaller cone?

r/R = H-h/H
r/6 = 12-h/12
r = 12-h/2

V = 1/3 r^2h
= 1/3 (12-h/2)^2 h
= 1/3 (144-h^2/4) h

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1. You are ok up to your last line:
[(12-h)/2]^2 = (144-24h+h^2)/4

so,

v = pi/12 (144h-24h^2+h^3))

dv/dh = pi/12(144-48h+3h^2)
dv/dh=0 when h=4,12

Naturally, v=0 when h=12, so
v(4)=64pi/3

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posted by Steve
2. V=64Pi/3
h=4,12

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