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January 27

Use Lagrange multipliers to solve the following

f(x,y) = x − 2 y^2 , g(x,y) = x^2 + y^2 − 1 = 0 and I got two test points, (-1/4, ±√15⁄4), which when plugged into f(x,y) gives the same value, but the book gives two different values. Where did I err? Thank you. 69.22.242.15 (talk) 11:23, 27 January 2017 (UTC)[reply]

You have not said what the problem is or why you chose those values or why you headed your query with Lagrange multipliers or what the values given in the book were. Dmcq (talk) 11:43, 27 January 2017 (UTC)[reply]

I'm studying Lagrange multipliers for the first time and this is a pretty straight forward problem with one function with two variables and one constraining condition. I just want to know if my answer is wrong. Since my test points vary only in the sign of the y-value. If you want to see my work, it goes like 1 = 2 λx and -4y=2λy, which gives λ=-2 and then the above. 69.22.242.15 (talk) 13:19, 27 January 2017 (UTC)[reply]

Sorry, forgot what they were. Anyway looking at the problem as far as I can see your application of the formulas is fine, but it doesn't find (1,0) which is a maximum. y=-0 is a solution of -4y=2λy. Dmcq (talk) 14:29, 27 January 2017 (UTC)[reply]

Solving your first order conditions for x and y in terms of

\lambda

gives y=0 and x=

{\frac {1}{2\lambda }}.

Using these in the constraint allows you to solve for

\lambda =\pm 1/2.

Using this in the earlier provisional solution for x gives x=1 and x=–1. Using these in the objective function f gives f= 1 and f=–1. Loraof (talk) 17:12, 27 January 2017 (UTC)[reply]

Blargh. Thank you both. 69.22.242.15 (talk) 17:28, 27 January 2017 (UTC)[reply]

Way to track best of x without being able to write it down

Example: Say I'm throwing free throws in basketball and want to track how many out of 10 I get in. I'm not interested in total, but I just want to hit a specific percentage of shots, say 60%, or 6 out of every 10. I want to be able to do this in my head, but I can't write it down or have anyone else track my shooting.

At first it would be easy: take ten shots, and I know exactly what my score is. But after 11 and onwards, what's the best way to only remember last 10. Can I do it like golf by using a par and using addition and subtraction to maintain a given score? What should that score be? Mingmingla (talk) 22:39, 27 January 2017 (UTC)[reply]

Not sure if this violates the "writing things down" part, but how about using some type of marker, like sticks (popsicle sticks would work). Vertical placement could mean it's good, and horizontal means it's bad. You would only have 10 sticks. After the 10th stick is placed, you start reusing the oldest stick, with each throw, placing it at the new end. You could quickly calculate the new running average of the last 10 throws and remember it, if higher than your previous max. Bending down to manage the sticks, on the ground, after each throw might be annoying, so ideally you would have a table you could put them on. StuRat (talk) 01:38, 28 January 2017 (UTC)[reply]

However, note that what you are doing is cherry-picking. Let's say you make each throw 50% of the time. Given that figure, the odds of making all 10 throws is 0.5¹⁰, or 1/1024, and that will tend to happen, on average, after 1024 tries. However, with your system, that's not even 1024x10 throws, but rather 1024+9 throws. Presumably that's well within your capability, so saying you once got 10 throws in a row is really a tribute to your perseverance, not skill. StuRat (talk) 04:34, 28 January 2017 (UTC)[reply]

I think the chances of having one or more runs of ten or more in a thousand goes is about 1/e. One could do it with ones fingers being up or down against a table and just changing them in sequence. as one circled round. The count would change if one has to move a finger. With a bit of practice I'm sure you could do that in your head just imagining the fingers or doors in a street or something like that. Dmcq (talk) 22:44, 30 January 2017 (UTC)[reply]

If the objective is to remember the last 10 numbers, then you have to remember them all, I fear. Mnemonics techniques may help, but at any point your memory will effectively store that information.

If you want to track the average success rate from the start without having to remember individual values, here you will find a so-called online algorithm to compute mean and variance of a series. For the mean, it is simple: (mean after shot n+1) = (n*(mean after shot n) + 1*(result of shot n+1))/(n+1). (Where "result of shot" is either 0 or 100% depending on whether the shot failed or succeeded) Tigraan^{Click here to contact me} 13:58, 31 January 2017 (UTC)[reply]

Thanks all. This is what I was afraid of. Mingmingla (talk) 02:17, 3 February 2017 (UTC)[reply]