Wikipedia:Reference desk/Archives/Mathematics/2009 May 4

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May 4

Application of Differentiation

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

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

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)

Penney's Game

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)

decimal to HEX

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 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 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)