Minimise the distance

Agnishom · 2013-03-25 22:46:18

<< Problem attached >>

Now this is what I do:

Construct lines between (1,5), (3,5), (4,9) and (1,9)
Now let us consider an arbitary point (1,y) and let the distance from (1,5) to (1,y) be r
Now we search for the value that would make ( 2^2 + r^2 ) + ((4 - r)^2 + 3^2) the minimum value
And the value of r which makes the above function to give its minimum is 2
So, is y = 5 + 2 the solution?

Last edited by Agnishom (2013-03-25 22:47:38)

bobbym · 2013-03-25 22:55:09

Hi;

I am getting 38 for a + b.

Agnishom · 2013-03-25 23:00:59

How to do it?

bobbym · 2013-03-25 23:10:59

For one thing always have a rough idea what the answer is. Sometimes you can have the exact answer before you even start with the the mathematics.

Agnishom · 2013-03-25 23:16:30

I believed it would be the midpoint

bobbym · 2013-03-25 23:21:19

A drawing also helps. Calculus will give an exact answer but geogebra got the answer to 15 places!

Agnishom · 2013-03-25 23:24:59

I used geogebra. But .... I couldn't find the exact

Okay, I'm ready for calculus

bobbym · 2013-03-25 23:29:18

y = 6.6 is the answer that geogebra comes up with.

This is the function to be minimized.

Can you do that?

Agnishom · 2013-03-26 00:00:20

So to minimise it we find its derivative f'
And then solve for f'(y) = 0
Because if that is 0 the slope is 0 and the horizontal line is formed and thats either the maxima or minima of the function

Am I correct?

bobbym · 2013-03-26 00:02:44

Yes, that is how calculus would do it. Again, I would urge you to graph the function first to have some idea where the minimum is.

My brother points out an nifty idea using vectors but I can not grasp it.

Agnishom · 2013-03-26 00:17:54

I haven't got an idea to find the derivative except using a CAS.
Actually, I am not fully acquinted with Calculus.

Is your brother on this forum?

bobbym · 2013-03-26 00:20:37

Yes, he is a member but he does not like forums.

If you can do it using a CAS that is okay too.

Agnishom · 2013-03-26 00:29:33

His username?

bobbym · 2013-03-26 00:42:08

jimmyR is his username.

Agnishom · 2013-03-26 01:02:40

Okay

The CAS didn't help me much. I haven't got a clue about equating that to 0 (by hand)

Last edited by Agnishom (2013-03-26 01:07:25)

bobbym · 2013-03-26 01:11:02

What CAS are you using?

Agnishom · 2013-03-26 01:15:19

MAXiMA,
I don't know much about it

bobbym · 2013-03-26 01:20:05

The standalone or the one bundled with geogebra.

Bob · 2013-03-26 01:27:48

OR

Reflect (3,5) in the line x=1 to give (-1,5)

Join this point to (4,9) to give the minimum distance.

EDIT: and the ratio is easily found by counting coordinate squares.

Bob

Agnishom · 2013-03-26 01:36:24

This sounds very interesting.
But:
1. That is not the answer
2. How do I know that that is the shortest distance?

Bob · 2013-03-26 01:46:50

hi Agnishom,

I have edited my hidden answer. Hopefully, you'll like the correction better.

The distance (3,5) to (1,y) and the distance (-1,5) to 1,y) are the same by the properties of reflection.

So minimise by drawing a straight line from (-1,5) to (4,9)

I am finding the site very slow at the moment ... not sure if this is general or not. It took 10 mins to make my last post, edit it and then correct the edit.

Hope what I have said is clear.

Bob

Agnishom · 2013-03-26 01:57:21

The site is okay for me

Yes, I got the idea Thanks a million

But I would be happier if you clarified it to be 33/5

Bob · 2013-03-26 08:54:32

Yes, sorry about that. My excuse is that I was being harassed by the slow re-load time on pages. It was driving me mad. Originally I just posted the method. Then I had the idea to work out the answer but hide it. But I was completely off track because I had been thinking that 'y' divided the line in the ratio a:b ie 2:3 so I posted a + b = 5.

Then I read the question properly (why don't I do this first???) and revised my answer. In my annoyance I thought 6 and 3/5 ... 6 5s are 35 add 3 makes 38. So a/b = 38/5. Big whoops. I realised that must be wrong because that wasn't what my diagram was telling me, but it took ages to get the edit screen to work so I could do 6 5s are 30 add 3. Pathetic excuse but there it is. I've admitted to the forum that I cannot do my five times table. I'll write it out 20 times.

Bob

ps. Load speed back to normal now.

Bob · 2013-03-26 09:00:07

FOR i = 1 TO 20
j = 1
REPEAT
ans = j*5
PRINT j;" x 5 = ";ans
j = j + 1
UNTIL j = 11
NEXT i

Bob

Agnishom · 2013-03-26 16:33:39

I have problems with my multiplication tables too. Is it very important?

Math Is Fun Forum

#1 2013-03-25 22:46:18

Minimise the distance

#2 2013-03-25 22:55:09

Re: Minimise the distance

#3 2013-03-25 23:00:59

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#4 2013-03-25 23:10:59

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#5 2013-03-25 23:16:30

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#6 2013-03-25 23:21:19

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#7 2013-03-25 23:24:59

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#8 2013-03-25 23:29:18

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#9 2013-03-26 00:00:20

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#10 2013-03-26 00:02:44

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#11 2013-03-26 00:17:54

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#12 2013-03-26 00:20:37

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#13 2013-03-26 00:29:33

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#14 2013-03-26 00:42:08

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#15 2013-03-26 01:02:40

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#16 2013-03-26 01:11:02

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#17 2013-03-26 01:15:19

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#18 2013-03-26 01:20:05

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#19 2013-03-26 01:27:48

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#20 2013-03-26 01:36:24

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#21 2013-03-26 01:46:50

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#22 2013-03-26 01:57:21

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#23 2013-03-26 08:54:32

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#24 2013-03-26 09:00:07

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#25 2013-03-26 16:33:39

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