 Subject: Sphere and cone
Date: Sun, 07 Dec 1997 03:43:36 -0500
From: mdt001@ibm.net

A question was posed to me, and I had some extreme difficulty obtaining an EXACT answer.

What is the height of the cone that has the maximum volume inscribed in a sphere of fixed volume?

Here is my try at it...PLEASE! see if it makes sense.

Volume of sphere, Vs = 4/3 * pi * Rs^3, where Rs is radius of the sphere. (volume fixed = radius fixed) Volume of cone, Vc = 1/3 * pi * h * Rc^2, where Rc is radius, h is height of the cone. (variable) Radius of cone as a function of height h, Rc(h) = Rs * sin(pi*h/2Rs).

As the height increases, the radius will grow until h=Rs, its maximum. It will then decrease from Rs to 2*Rs. Start at h=0, the sin term disappears, following the shape of the sphere, Rc(h) increases until h=Rs, and the sin term becomes, sin(pi/2)=1, so Rc=Rs=h. Now it decreases, until the cone diminishes at the other pole.

Substituting Rc into Vc, we find the volume of all the cones at height h, Vc(h) = 1/3 * pi * h * (Rs * sin(pi*h/2Rs) )^2.

Now, differentiate Vc(h) with respect to h to find critical points. dVc(h)/dh = h/3 * Rs^3 * sin(pi*h/2Rs) * cos(pi*h/2Rs) + pi/3 * Rs^2 * (sin(pi*h/2Rs))^2

Unfortunately, dV/dh=0 is quite difficult to solve for non-trivial solutions. If h=0, no problem, but 0>h>Rs is rough. It is a linear function, but its actual formula eludes me. The closest I can come is h=1.1692140934*Rs, thanks to my HP, but I am sure there MUST be an EXACT answer for this problem. |Reply| |Previous| |Next| |Down| |Exchange index| |Contents| |Store|