something MATH.......

[lovelove]

I know this is the wrong forum, but please help me.
It is about a simple looking math problem.

A = 2xy / (x^2 + y^2)
x>0, y>0
What is the maximum of 'A'?

I'm stuck with this. I'd appreciate anyone who can show the solution.


Winter vacation has ended, and the university enterence test is at November 2013.... I sadly have terribly short time for Go..

[Shaddy]

I know this is the wrong forum, but please help me.
It is about a simple looking math problem.

A = 2xy / (x^2 + y^2)
x>0, y>0
What is the maximum of 'A'?

I'm stuck with this. I'd appreciate anyone who can show the solution.


Winter vacation has ended, and the university enterence test is at November 2013.... I sadly have terribly short time for Go..

It is 1. There is a common inequality (called arithmetic-geometric mean inequality, but nevermind that) that says that

x^2 + y^2 >= 2xy.

to prove it, you can write x^2 + y^2 - 2xy = (x-y)^2 >= 0

[lovelove]

I know this is the wrong forum, but please help me.
It is about a simple looking math problem.

A = 2xy / (x^2 + y^2)
x>0, y>0
What is the maximum of 'A'?

I'm stuck with this. I'd appreciate anyone who can show the solution.


Winter vacation has ended, and the university enterence test is at November 2013.... I sadly have terribly short time for Go..

It is 1. There is a common inequality (called arithmetic-geometric mean inequality, but nevermind that) that says that

x^2 + y^2 >= 2xy.

to prove it, you can write x^2 + y^2 - 2xy = (x-y)^2 >= 0

[mitsun]

For an alternative geometric solution, you could draw a circle and transform to polar coordinates:
x = r sin(t), y = r cos(t), 2xy / (x^2 + r^2) = (some trig manipulation) = sin(2t)

[TheBigH]

For an alternative geometric solution, you could draw a circle and transform to polar coordinates:
x = r sin(t), y = r cos(t), 2xy / (x^2 + r^2) = (some trig manipulation) = sin(2t)

Ooh, that's elegant. Nice idea!

[drmwc]

If you've done calculus, a possible thrid approach is to take the partial derivative of A w.r.t. x or y, and set it to 0.

This a lot less elegant than the solutions posted so far, though.

Specifically:

dA/Dy=(2x^3-2xy^2)/(x^2+y^2)^2.

Setting this to 0, and using x>0 and y>0, gets x^2=y^2. Hence x=y at the maximum, where A=1. (There's a little more work to do to show this a maximum not a minimum.)

[speedchase]

If you've done calculus, a possible thrid approach is to take the partial derivative of A w.r.t. x or y, and set it to 0.

This a lot less elegant than the solutions posted so far, though.

Specifically:

dA/Dy=(2x^3-2xy^2)/(x^2+y^2)^2.

Setting this to 0, and using x>0 and y>0, gets x^2=y^2. Hence x=y at the maximum, where A=1. (There's a little more work to do to show this a maximum not a minimum.)

You also have to show that +/- infinity is greater

[lightvector]

If you have the right intuition and bits of knowledge, it's possible to quite quickly see the solution to this problem in your head, without even having to write anything down.

The denominator, x^2+y^2, is the squared length of the vector (x,y). And for any particular value of the denominator, that is, for any particular fixed length, how do you choose a vector (x,y) of that length that maximizes the numerator, 2xy? Geometrically, what points on a circle around the origin maximize xy? It should be intuitively obvious that the maximum occurs when x = y.

And when x = y, the whole expression simplifies into 2xx/(x^2+x^2) = 2x^2/(2x^2) = 1, so the maximum is 1 and does not depend on the length of the vector, aside from it needing to be nonzero.

(This is mostly the same as the polar coordinate solution).

[lovelove]

Real thanks for everyone who replied (and will reply...) to my post.
And... let me commit my second abuse to this forum.

The (math) problem, below in a italic font, that I'm again in trouble with wants me to find the truth values of propositions about matrix. The original problem is written in Korean, so I may have used some improper mathematical expressions in English.

'A' is a 2x2 matrix, 'O' is a 2x2 zero matrix, 'n' ∈ a set of all natural numbers

(1)
A^n = O → A^2 = O

(2)
A 2x2 matrix 'X' that 'X^2 = A' always exists regardless of the value of 'A'. (Whatever value 'A' has, there must be some kind of a 2x2 matrix 'X' that 'X^2 = A')

(3)
'A' that 'A^2 = ( 1 2 )' exists.
____________( 3 4 )

I'd again, of course, appreciate anyone who can show the solution.

[Bill Spight]

Real thanks for everyone who replied (and will reply...) to my post.
And... let me commit my second abuse to this forum.

The (math) problem, below in a italic font, that I'm again in trouble with wants me to find the truth values of propositions about matrix. The original problem is written in Korean, so I may have used some improper mathematical expressions in English.

'A' is a 2x2 matrix, 'O' is a 2x2 zero matrix, 'n' ∈ a set of all natural numbers

(1)
A^n = O → A^2 = O

(2)
A 2x2 matrix 'X' that 'X^2 = A' always exists regardless of the value of 'A'. (Whatever value 'A' has, there must be some kind of a 2x2 matrix 'X' that 'X^2 = A')

(3)
'A' that 'A^2 = ( 1 2 )' exists.
____________( 3 4 )

I'd again, of course, appreciate anyone who can show the solution.

To get you started:

Code: Select all

     A = (a11 a12)
         (a21 a22)

What is A^2?

[lovelove]

Code: Select all

     A = (a11 a12)
         (a21 a22)

What is A^2?

Code: Select all

A = (a b)
    (c d)
A^2 = (a^2+bc ab+bd)
      (ac+cd d^2+ab)

[Bill Spight]

Code: Select all

     A = (a11 a12)
         (a21 a22)

What is A^2?

Code: Select all

A = (a b)
    (c d)
A^2 = (a^2+bc ab+bd)
      (ac+cd d^2+ab)

You have an oops!

OK. What must be true of a, b, c, and d for A^2 to be 0?

[lovelove]

Code: Select all

     A = (a11 a12)
         (a21 a22)

What is A^2?

Code: Select all

A = (a b)
    (c d)
A^2 = (a^2+bc ab+bd)
      (ac+cd d^2+ab)

You have an oops!

OK. What must be true of a, b, c, and d for A^2 to be 0?

a=b=c=d=0 or something like a=1, b=1, c=-1, d=-1 (some cases when ad-bc = 0)

[Bill Spight]

Code: Select all

A = (a b)
    (c d)
A^2 = (a^2+bc ab+bd)
      (ac+cd d^2+ab)

You have an oops!

OK. What must be true of a, b, c, and d for A^2 to be 0?

a=b=c=d=0 or something like a=1, b=1, c=-1, d=-1 (some cases when ad-bc = 0)

Do you see the oops?

Make a general statement about the conditions for A^2 = 0.

I am going to bed now.

[Bill Spight]

A <> 0, A^2 = 0 , where A is a 2x2 matrix.

Write A using 2 real variables.

----

Given: A is a 2x2 matrix.
Prove: A^4 = 0 -> A^2 = 0.