Cone in a Ball

Cone with maximal volume inscribed in a hollow ball

You have a hollow plastic ball that opens into two halves.
(Note: The plastic spheres can be purchased at Hobby Lobby or Michaels. We found four different sizes, varying in price from 99¢ to $1.49.)

Task.
Design and make a cone that can be inscribed in the ball. The cone should have the biggest volume possible.

Solution

See the illustration, which shows a 2-D cross-section of the sphere and cone.

Variables

cos(x)=s/(2r) (why? Look at triangle ABC.) So
s = 2r*cos(x)
sin(x)=b/s (look at triangle ABE), so
b=s*sin(x) = 2r*cos(x)*sin(x)
Also cos(x) = h/s, so
h=s*cos(x) = 2r*cos²(x)
V = 1/3*(area of base)*height = 1/3*(π*b²)*h
= 1/3*π*(2r*cos(x) *sin(x))²*2r*cos²(x)
= 8/3*π*r^3*cos⁴(x) *sin²(x)
Now let t = cos(x), and let c = 8/3*π*r³
Do you know the identity sin²(x) + cos²(x) = 1?
We use the fact that sin²(x) = 1- cos²(x)
V = c*t⁴*(1-t²)
= c*(t⁴-t⁶)
Now we take the derivative with respect to t, and set it equal to zero, to find where V is maximum. We can do this by hand or with a calculator.

With a calculator,
enter Y₁=X^4-X^6
Under MATH, choose SOLVER. Enter
nDeriv(Y₁,X,X)=0
Enter a guess between 0 and 1.
X=.6
Press ALPHA SOLVE
You will see X=.816495...
From the home screen, enter
cos^-1(X)
You will see
35.264....

By hand,
dV/dt = c*(4*t³-6t⁵)
But 0 < t < 1. So dV/dt = 0 when 4 - 6t² = 0, which gives,
t = √(2/3)
And cos^(-1) (√(2/3)) ≈ 35.26438968 degrees
(≈ means "approximately equals")
So 2x ≈ 70.5 degrees.

What next?
We know x (see illustration above), and we measured r. We need to design the 2-D plan for the cone. It will be a sector of a circle with radius s. (Do you see this?) What we need to know is the value of angle A (see illustration below).

s=2r*cos(x), so we can compute s. (Of course we have to compute the radius r first.)
circumference of cone = 2*π*b = 2*π*s*sin(x) (we can compute this)
circumference of circle with radius s = 2*π*s.
We need angle A. Notice

Do you see why?
sin(x)* 360° ≈ 207.846° ≈ 208°.

Now we can design our cone.
Draw and cut out the pattern, tape it, and see if it fits in your sphere!

Two questions:
1. What is the ratio of the height of the cone to the diameter of the ball? With just a bit of algebra, you will find that it is 2/3!
2. What percentage of the volume of the ball does the cone take up? (You can estimate this by filling both with rice and pouring it into a graduated beaker, or you can do it with math.)

Another task. Design a cone that fits in your sphere with maximal surface area. There are two cases: include the base of the cone in the surface area, or leave the cone open, without a base. Are the two cones identical?