Out of a circular piece of paper, you wish to form a cone cup, so you cut out a circle wedge (with its extreme at the circle center) and join the resulting straight sides, forming a conical cup.
What size should the wedge be, to maximize the capacity of the cone?
(In reply to
Very rough approximation by Erik O.)
I'm sure what you maximized was r^2 * h, rather than the area of the triangle. First of all, that's what makes sense, and second, as the following table shows, .58 for h does not maximize the area of the triangle (h*r/2):
h r h*r/2 h*r^2
0.50 0.8660 0.2165 0.3750
0.51 0.8602 0.2193 0.3773
0.52 0.8542 0.2221 0.3794
0.53 0.8480 0.2247 0.3811
0.54 0.8417 0.2272 0.3825
0.55 0.8352 0.2297 0.3836
0.56 0.8285 0.2320 0.3844
0.57 0.8216 0.2342 0.3848
0.58 0.8146 0.2362 0.3849
0.59 0.8074 0.2382 0.3846
0.60 0.8000 0.2400 0.3840
0.61 0.7924 0.2417 0.3830
0.62 0.7846 0.2432 0.3817
0.63 0.7766 0.2446 0.3800
0.64 0.7684 0.2459 0.3779
0.65 0.7599 0.2470 0.3754
0.66 0.7513 0.2479 0.3725
0.67 0.7424 0.2487 0.3692
0.68 0.7332 0.2493 0.3656
0.69 0.7238 0.2497 0.3615
0.70 0.7141 0.2499 0.3570
0.71 0.7042 0.2500 0.3521
0.72 0.6940 0.2498 0.3468
0.73 0.6834 0.2495 0.3410
0.74 0.6726 0.2489 0.3348
0.75 0.6614 0.2480 0.3281
0.76 0.6499 0.2470 0.3210
0.77 0.6380 0.2456 0.3135
0.78 0.6258 0.2441 0.3054
0.79 0.6131 0.2422 0.2970
0.80 0.6000 0.2400 0.2880
It looks like h=.71 would come closest to maximizing the area of the right triangle; but that's not what we want to do.
The formula relating h and r would be r^2 = 1h^2.
The volume of the cone then becomes pi*h*(1h^2)/3.
Differentiating this and setting to zero we get
13*h^2 = 0
so that h=1/sqrt(3) and r=sqrt(2/3).
The circumference of the base is then 2*pi*sqrt(2/3), while the circumference of the original disk was 2*pi, so the fraction (1sqrt(2/3)) of the original circumference is to be removed. This fraction of 360 degrees is 66.06123086601863 degrees.
Edit: Corrected column headings.
Edited on April 1, 2005, 7:21 pm

Posted by Charlie
on 20050401 19:16:30 