b =

radius of base of cone (we don't know this)


r =

radius of ball (we measure this)


h =

height of cone (we don't know this)


X =

angle (also unknown; see diagram; if we find the value for
X, we can find values for all other variables)


(Notice
that π and r are constants.)
The minimum volume is reached for some value of X between 0 and 45 degrees. For
this value, the derivative dV_{c}/dX = 0.
Computation with the TI83/84 Plus.
Set MODE to degrees.^{ }
Define,^{ }
\Y_{1}=tan(2X)/tan(X)^{3}
(Or you may set \Y_{1}=(π/3)*R^{3}*tan(2X)/tan(X)^{3}. If you do
this,^{ }
\Y_{2}=nDeriv(Y_{1},X,X) be sure to
store the golf ball's radius in R.)^{ }
Set
Solver to^{ }
Y_{2}=0^{ }
X=30 (for example)^{ }
bound={0,45}^{ }
and ALPHA SOLVE.^{ }
You get X=35.26438965271.^{ }
From above, the values of b and h, in terms of r and X, are^{ }
b = r/tan(X) and h = r*tan(2*X)/tan(X)^{ }
The volume of the ball is V_{b} = 4/3*π*r^{3}, and thus^{
}
V_{b}/V_{c} = 4*tan(X)^{3}/tan(2*X).^{ }
Ratio of the volume of the ball to the volume of the cone:^{ }
4tan(X)^{3}/tan(2X) ENTER returns .5. So the ball occupies exactly one
half of the cone!^{ }
Design and construction.
Measure the diameter of the golf ball and compute its radius r. Follow the
method of designing a cone of given dimensions described in Baggett &
Ehrenfeucht, 2001. It is also included here.