Solution to the Cone Inscribed in the Sphere Problem© 2001 by Karl Hahn |
The problem is choose the dimensions of a cone, height h, and base radius r, in such a way that in can be inscribed in a sphere of radius, R, and maximize the volume of the cone. You have to use some trigonometry to solve this one. The figure on the right shows a cross section of the cone inscribed in the sphere. The point, O, is the center of the sphere. The point, A, is the apex of the cone. The green isosceles triangle is the outline of where the cone intersects the midplane of the sphere. You also have the line segment
__ AC = |
axis of the cone |
__ OD |
bisects angle |
AOB |
__ OD |
intersects |
__ AB |
at right angles |
__ AD = R cos(θ)Since triangle, AOB, is isosceles, you also get length
__ AB = 2R cos(θ)which is the hypotenuse of the right triangle, ABC. Using the "sine equals opposite over hypotenuse" rule, you can establish the length
__ BC = r = 2R cos(θ)sin(θ)If you check the trig identity page, you find that 2 cos(θ)sin(θ) = sin(2θ) (equation 7.1b-6b).
__ BC = r = R sin(2θ)Likewise, using the "cosine equals adjacent over hypotenuse" rule, you can establish
__ AC = h = 2R cos^{2}(θ)Also from the trig identity page (equation 7.1b-6a), you can easily get 2 cos^{2}(θ) = 1 + cos(2θ). So:
__ AC = h = R (1 + cos(2θ))
The formula for the volume of a cone is
1 V =Substituting from the above for r and h, this becomesπr^{2}h 3
1 V =In most min-max problems where you have two or more independent variables (here we have independent variables r and h) you try to solve for one in terms of the other. But here you can begin to see another alternative. We have solved for both r and h in terms of a third, related variable, θ. And we have expressed the dependent variable, volume, V, in terms of θ alone. So now we can find the value of θ that maximizes V. To do the problem that way, you would find the derivative, dV/dθ, and set it to zero.πR^{3} sin^{2}(2θ)(1 + cos(2θ)) 3
But taking the derivative of the product of those trig functions is messy. There is an even easier way. Notice from the trig identity, sin^{2}(2θ) = 1 - cos^{2}(2θ) the volume equation becomes
1 V =We can get rid of all the trig functions from this simply by substituting a new variable, u = cos(2θ).πR^{3} (1 - cos^{2}(2θ)) (1 + cos(2θ)) 3
1 V = |
1 |
dV 1Solving the quadratic for u using the quadratic formula:= 0 =πR^{3} (1 - 2u - 3u^{2}) du 3
______ 2 ± √4 + 12 u = |
or | 1 |
h = R (1 + cos(2θ)) = R(1 + u)So either
h = 0 |
or | 4 h = |
______ r = √1 - u^{2} R
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