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
r
radius of the sphere (you measure this; be sure to measure the interior radius).
b
radius of base of cone
s
slant height of cone
x
one half the angle at the apex of the cone
V
Volume of cone
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*cos^{2}(x)^{ }_{ }
V = 1/3*(area of base)*height = 1/3*(π*b^{2})*h^{ }_{ }
= 1/3*π*(2r*cos(x) *sin(x))^{2}*2r*cos^{2}(x)^{ }_{ }
= 8/3*π*r^3*cos^{4}(x) *sin^{2}(x)^{ }_{ }
Now let t = cos(x), and let c = 8/3*π*r^{3}^{ }_{ }
Do you know the identity sin^{2}(x) + cos^{2}(x) = 1? ^{ }_{ }
We use the fact that sin^{2}(x) = 1- cos^{2}(x)^{ }_{ }
V = c*t^{4}*(1-t^{2})^{ }_{ }
= c*(t^{4}-t^{6})^{ }_{ }
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 Y_{1}=X^4-X^6
Under MATH, choose SOLVER. Enter
nDeriv(Y_{1},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*t^{3}-6t^{5}) ^{ }_{ }
But 0 < t < 1. So dV/dt = 0 when 4 - 6t^{2} = 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?