Question

Find \(\text{min} \left[ (x_1 - x_2)^2 + (12 + \sqrt{1-{x_1}^2 } - \sqrt{4 x_2} )^2 \right] \) for all \(x_1 , x_2 \in R\)

Answer 1

@ganeshie8 @mathmate @hartnn @perl

Answer 2

I'm myself confused. Not sure whether Calculus will do it or not! But very sure, this has got some relation with the Conic Sections. But again, not sure, how!

Answer 3

great work :P :D

Answer 4

wolfram gives messy numbers hmm
http://www.wolframalpha.com/input/?i=minimum+%28a-b%29%5E2+%2B+%2812+%2B+sqrt%281-a%5E2%29+-+sqrt%284b%29%29%5E2

Answer 5

I'm thinking \[x_{1} = 1\] so \[\sqrt{1-x_{1}^2} = 0\] and you get something that is easier to plot.

Answer 6

you want to minimize
f(x,y)

Answer 7

On that plot, you'll see a min value

Answer 8

We calculate the Hessian

Answer 9

it should come out the same as what ganeshie8 put into wolfram

Answer 10

min {(x1-x2)^2+(12+sqrt(1-x1^2)-sqrt(4*x2))^2}

= 1/3 (433-7775/(140112 sqrt(2919)-7538831)^(1/3)+(140112 sqrt(2919)-7538831)^(1/3))

Answer 11

which is approximately 72.41145427

Answer 12

\((x_1-x_2)^2\geq 0\)
\((12+\sqrt{1-x_1^2}-\sqrt{4x_2})^2\geq0\)
-----------------------------------------
the sum \(\geq\) 0
so, the minimum if there is some is =0

Answer 13

I don't think that there are any possible values for the function = 0 .

Answer 14

\(x_1-x_2\geq \pm (12+\sqrt{1-x_1^2}-\sqrt{4x_2})\)

Answer 15

First find and classify the critical points of
f(x,y) = [ (x - y)^2 + (12 + sqrt(1-x^2) - sqrt(4y) )^2 ]

Solve the system of equations
fx = 0
fy = 0

Answer 16

Finding the critical points... hmm, but how?

Answer 17

fx = partial f(x,y) / dx
fy = partial f(x,y) /dy

Answer 18

Paul explains it here
http://tutorial.math.lamar.edu/Classes/CalcIII/RelativeExtrema.aspx

Answer 19

Okay so here is what I can think about :

Let : \( 12 + \sqrt{1 - {x_1}^2 } = A \)
\((A-12)^2 = 1 - {x_1} ^2\)

And say, \(\sqrt{4 x_2} = B\)

\(B^2= 4 x_2 \)
Which is infact an equation of parabola.

Answer 20

Trust me, it is an objective question and as per the time given for each question (2-3 minutes), I don't think that doing this question with calculus will be appropriate. It has really got something to do with Conics.

@Kainui @vishweshshrimali5

Answer 21

Are you allowed to use some graphing tool?

Answer 22

Unfortunately, no @aleroth . Only allowed to use ... one pen, paper.. .and that's it :(

Answer 23

ok, so you are looking for a pre-calculus argument

Answer 24

Is this question listed only under "conic sections" or some specific chapter under it? I haven't really started with conics but one geometric interpretation, which I'm not sure is a good start, would be considering these two points:\[P(x_1, 12 + \sqrt{1 - x^2})\]\[Q(x_2, 2\sqrt{x_2})\]...and then minimizing the distance between them.

Answer 25

Now if we do find some nice way to minimize distances other than using the distance formula, then this problem would be a piece of cake. But the reason why this problem is a good one is that I don't think there is. :P

Answer 26

Exactly! That is what I was trying to do. Haha! kya karun yaar mujhe buddha mil gaya... :/ This question is seriously... :'( Mind boggling :P

Answer 27

Very interesting! so you're basically trying to find the minimum distance between a circle and a parabola :

\(x^2+(y-12)^2 = 1\)
\(y^2=4x\)

Answer 28

those equations follow trivially from PK's previous reply ^

Answer 29

Oh wow, I somehow skipped mathslover's reply. That looks like it.

Answer 30

Is there an easy way to find the distance between conics ? XD

Answer 31

I cheated a little... https://answers.yahoo.com/question/index?qid=20120203010806AAxz86W

Answer 32

I'm going to exile... for 100 years.. LOL!

Answer 33

it would be nice to see if you get the same answer as the calculus solution :)

Answer 34

and how did wolfram get the answer so quickly (did they use calculus)?

Answer 35

A trick I learned in calculus 1 for minimzation problems is that you can look for things that don't affect the minimum point and remove them, such as square roots or constants. Not sure if that'll help here or not, but probably. A good example would be: \[\Large f(x) = \sqrt{1-x^2} \\ \Large g(x) = 1-x^2\] both have the same minimum for x. Another example: \[\Large f(x)= x^2+7 \\ \Large g(x) = x^2\] Also have the same minimum for x.

So some things to think about to quickly throw away some stuff maybe?

Answer 36

*Just be careful when using this rule ;)

Answer 37

this question made my day! working it is a simple matter of finding the normal and yes it can be done in less than 2 minutes XD

Answer 38

Another thing (didn't read so might have been said) x2 has to be greater than 0 otherwise you get a negative under the square root.

Answer 39

Using this widget I get the critical point (x,y) is (-sqrt(5)/5, 4)
so the minimum is f( -sqrt(5)/5, 4)

http://www.wolframalpha.com/widgets/view.jsp?id=597dd91e6b177f393877f173f231593f

Answer 40

Woah this geometric interpretation seems like the way to go, but I don't think I would have gotten there on my own haha.

Answer 41

yes

Answer 42

this is reallly a very seductive trick! but it only works in this special case and im sure im gona forget this after few days

Answer 43

the author cooked up the problem by taking two conics and squaring the distance between them - what cunning mind haha!

Answer 44

Can someone help me draw this picture, I'm kind of stuck: