How to solve this problem:
Also, most people would use trigonometry, but is there a way to use derivative to solve this too?
How to solve this problem:
Also, most people would use trigonometry, but is there a way to use derivative to solve this too?
Above, $O$ is the centre of the sphere, $x$ is its radius. Writing the area of trapezoid $ABCD$ as a sum of three triangles, we can easily solve for $x$ to get:
$ x=\frac{1200\sqrt{3}}{54}=38.490017\ldots $
No trig or calculus needed, only geometry.
Here is another solution appealing to the vector line equation for the cone edge, and the required distance r from the sphere center O to the cone edge.
Using the drawing above, which places one cone's base-center on the x-axis, and the sphere's center on the z-axis, and projecting the problem as a 2D problem on the x-z plane, the shortest line segment LS from the sphere center to the cone runs perpendicular to the cone edge. There are 2 constraints we can use to solve for r: The sphere center is at z coordinate 120-r, and the distance of the sphere-center [ 0, 120-r] from the cone edge is also equal to r.
In order to find the distance of a point from a line, I used the "orientation-location" vector line equation p • o = L to represent the cone edge, where p is any point [ x, y ] on the cone edge, o is the orientation of the line (unit vector pointing perpendicular to line), and L is the "location" of the line (signed distance of line from origin where o defines positive distance).
The signed proximity of a point p1 from a line is:
signed_proximity (p1) = p1 • o - L
Now just apply the vectorized line equation to the problem specifics:
Line equation of cone edge: o = [ 12/13, -5/13 ] (points right and slightly down)
L = [ 100/sqrt(3), 120 ] • o (use any known point on line to compute L) = 7.1400
Signed proximity (should be -r based on direction or o):
-r = [ 0, 120-r ] • [ 12/13, -5/13 ] - 7.1400 (solve for r) r = 13/18 [ 120*(5/13) + 7.1400 ] = 38.49