All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes
Cutting Contrives Conical Cup (Posted on 2005-04-01) Difficulty: 3 of 5
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?

See The Solution Submitted by Old Original Oskar!    
Rating: 4.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution re: Very rough approximation | Comment 4 of 18 |
(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 = 1-h^2.

The volume of the cone then becomes pi*h*(1-h^2)/3.

Differentiating this and setting to zero we get

1-3*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 (1-sqrt(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 2005-04-01 19:16:30

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (1)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information