Wikipedia:Reference desk/Archives/Mathematics/2009 May 4#Penney.27s Game

A piece of wire of length k cm is bent to from a sector. Find the maximum area of the sector.

There has given some equations, please use these equations to slove my question.

Let r be the radian, A be the area, y be the arcs.

k=2r+y.....(1)

y=2πr* x/360.....(2)

A=πr^2 (x/360) ....(3) --Dansonncf (talk) 11:10, 4 May 2009 (UTC)

:Okay, so the wire is bent to form a sector, with two straight edges of length r each, and a circular arc with radius r and length k-2r. The angle of this arc is (k-2r) / r radians, so the area enclosed by the wire is A = (k-2r)r / 2 = (kr / 2) - r². When r = 0, then A is 0 and when r = k / 2 then A is also 0. Somewhere in between these limis is a value of r that maximises A = (kr / 2) - r². At this maximum, dA/dr will be 0 ... Gandalf61 (talk) 12:39, 4 May 2009 (UTC)

:: Wrong. Radius can not be zero, because then the whole wire must be twisted infinitely many times in the single point. Assuming the largest possible angle of a sector is 2π, we have k = 2πr + 2r, which gives the lower bound of r ≥ k/(2(π+1)). --CiaPan (talk) 13:50, 4 May 2009 (UTC)

:::Picky, picky. I didn't want to confuse the questioner by introducing physically realistic limits at the lower end. Nevertheless, the function that you want to maximise is (kr / 2) - r², and if you realise that this quadratic has roots at r=0 and r=k/2 then you can use the symmetry of the quadratic to reach a non-calculus solution, as Michael Hardy suggests below. Gandalf61 (talk) 14:55, 4 May 2009 (UTC)

::::Indeed, the objection that the radius can't be zero is totally irrelevant. One first solves the mathematical problem, and then discusses the solution he's found. Here, imposing an a priori range on the radius would only turn in unnecessary complications. See e.g. Gandalf's argument: The area of a sector is bilinear as a function of r and l (the length of the arc). Here, the constraint is affine in r and l. Therefore, in the constraint, the area is some quadratic polynomial of r. For r=k/2 the area is zero, because the sector has angle zero. For r going to 0 the area (with multiplicity) must also go to 0 (the length is bounded). Now, a parabola is symmetric, and you have your maximizing radius k/4, without writing any formula. As you see, r=0 is mathematically meaningful. (PS: sorry for inserting myself, I planned a shorter post)--pma (talk) 15:46, 4 May 2009 (UTC)

The length of the circular arc is k − 2r. The circumference of the whole circle is 2πr. The fraction of the circle's area is therefore (k − 2r)/(2πr). The area of the whole circle is πr². The area of the sector is therefore

: $A\; =\; \backslash frac\{k\; -\; 2r\}\{2\backslash pi\; r\}\backslash cdot\; \backslash pi\; r^2\; =\; \backslash frac\{(k-2r)r\}\{2\}.$

You don't need differentiation to find the value of r that maximizes this, since is just a quadratic polynomial in r, so it's just a matter of finding the vertex of a parabola by completing the square. You already know how to do that before you take calculus. But you can do it by differentiation. Michael Hardy (talk) 14:40, 4 May 2009 (UTC)

Okay, just read Penney's game, and it makes exactly zero sense to me.

Probability of getting a specified particular result in a fair coin toss: 1/2. Probability of getting any specified sequence of three: 1/2^3 = 1/8. Probability of getting any one specified sequence before any one other specified sequence: 1/2. Each sequence has an equal probability of occurring. I have no idea what the linked Java simulation is doing, but I coded up a trial (here) and the result is a pretty clear 1/2.

Can someone explain what this article is talking about?

Thanks a lot, Aseld ^talk 19:07, 4 May 2009 (UTC)

: Suppose A picks HHH, and B picks (according to the strategy table) THH. The only way A can possibly win is if the first tosses of the game are HHH, which happens 1 time in 8. If that doesn't happen, there's no way for HHH to come up without THH coming up first, which means B wins. The other strategies work similarly. -- Coneslayer (talk) 19:15, 4 May 2009 (UTC)

:: Okay, got it, thank you. I wasn't visualising it properly - for some reason I was thinking of independently-generated groups of three. --Aseld ^talk 19:18, 4 May 2009 (UTC)

: (Edit conflict—you got it!) PS: In looking at your code, I think you're flipping the coin in sets of three flips, checking for a match, three more flips, checking for a match, etc.—that's not how the game is played. It's one flip at a time, checking for a match after each flip (using the three most recent). I think that's your misunderstanding. -- Coneslayer (talk) 19:19, 4 May 2009 (UTC)

What is the largest decimal number that will spell an English word in HEX? Is it 251636973? 65.121.141.34 (talk) 20:09, 4 May 2009 (UTC)

:There's nothing longer in [http://www.visca.com/regexdict/ this regex dictionary] (unless of course you accept cheats like o=0, l=1). Algebraist 20:25, 4 May 2009 (UTC)

:(edit conflict) That's probably the largest decimal number that spells an uncapitalized, non-hyphenated word. There are larger examples, if you broaden your vocabulary to include proper nouns or hyphenated words; for example, 4206546606 or [http://www.answers.com/topic/face-bedded-edge-bedded-1 1077210439149]. —Bkell (talk) 20:29, 4 May 2009 (UTC)

::Allowing Algebraist's "cheats," you can get 16612964617902. —Bkell (talk) 20:35, 4 May 2009 (UTC)