Please help me with this math problem.

[StatsManD]

I tried other message boards, but no one could prove this on them.


given:

(x^2+y^2)/(x^2 + 2*x*y - y^2) >= c/(x + y) , and c <= 1, and x+y <=1, x>=0 and y>=0.
Prove that when c = 0 the area shaded is 1/2, and as c gets larger the area shaded decreases monotonically.


Then prove that this statement also holds true when you interchange x and y. So same equation but this time switch x and y.

 

[KLin]

42

 

[Epic Fail]

registered just to ask for hw help? :roll:

 

[chuckywang]

What shaded area?

 

[StatsManD]

The area represented by the inequality, is the shaded area. This is an algebra 1 problem people.

 

[Kirby]

Originally posted by: yamadakun
registered just to ask for hw help? :roll:

But he'll stay for the YAGTs 😛

 

[StatsManD]

I didn't register just for this. This was one reason though.

 

[StatsManD]

This is a pre algebra problem. Most people here could do it, so someone help me.

 

[beemercer]

Originally posted by: StatsManD
This is a pre algebra problem. Most people here could do it, so someone help me.

Sure we could do it, but that requires thought.

Also am I reading it right, you have 42 posts and you joined today 😕.

 

[txrandom]

graph it out.

 

[StatsManD]

42 isn't so bad is it. Just have lots of things to say, and lots of questions to ask.

 

[StatsManD]

Originally posted by: txrandom
graph it out.

I would like to see you graph this by hand.

Graphing this problem in it self is almost impossible and very difficult.

 

[StatsManD]

This problem comes from a high level math class.

 

[beemercer]

Originally posted by: StatsManD
42 isn't so bad is it. Just have lots of things to say, and lots of questions to ask.

You're at 45 per day now, just keep it up and you'll pass me in 9 days :frown:.

 

[StatsManD]

Where do you get those smilies. I just see three when I hit reply.

 

[tfinch2]

Originally posted by: StatsManD
The area represented by the inequality, is the shaded area. This is an algebra 1 problem people.

Originally posted by: StatsManD
This is a pre algebra problem. Most people here could do it, so someone help me.

Originally posted by: StatsManD
This problem comes from a high level math class.

😕

DO YOUR OWN HOMEWORK

 

[2Xtreme21]

I love how your rationale for us to help you is "this comes from an Algebra 1 course so you all should be able to do it." If it comes from an Algebra 1 course, which you're probably in, YOU should be able to do it.

RTFM. Math is easy if you put effort in to it.

 

[StatsManD]

Originally posted by: tfinch2

Originally posted by: StatsManD
The area represented by the inequality, is the shaded area. This is an algebra 1 problem people.

Originally posted by: StatsManD
This is a pre algebra problem. Most people here could do it, so someone help me.

Originally posted by: StatsManD
This problem comes from a high level math class.

😕

DO YOUR OWN HOMEWORK

I thought if I lied about it being easy more people would attempt it. I know I should do it my self, but I can't figure it out.

 

[beemercer]

Originally posted by: StatsManD
Where do you get those smilies. I just see three when I hit reply.

Well if you use the full "Replying to Message" separate window they should be under "Select Emotion", or you memorize them.

🙂 😉 🙁 😱 :disgust: 😀 :| :Q 😛 😎 :frown: 😕 (definitely the best) :heart: :brokenheart: :beer: :music: :wine: :lips: :camera: :gift:

:evil: :clock: :light: :sun: :moon: :roll: :laugh: :shocked: :thumbsup: :thumbsdown: .

 

[2Xtreme21]

Yeah that makes us wanna help you.

 

[Vertimus]

Nobody's going to solve your problem if you don't state the problem correctly. When c=0, the area of the shaded region is only 1/2 if you restrict yourself to the first quardrant, which you didn't specify.

 

[Rubycon]

Originally posted by: StatsManD
Where do you get those smilies. I just see three when I hit reply.

x = a sec(t), y = b tan(t) :laugh:

 

[StatsManD]

Originally posted by: Vertimus
Nobody's going to solve your problem if you don't state the problem correctly. When c=0, the area of the shaded region is only 1/2 if you restrict yourself to the first quardrant, which you didn't specify.

My bad, forgot that little tidbit. Yes we must also assume x>=0 and y>=0.

 

[fishmonger12]

Originally posted by: StatsManD

Originally posted by: txrandom
graph it out.

I would like to see you graph this by hand.

Graphing this problem in it self is almost impossible and very difficult.

ROFL. That is all.

 

[StatsManD]

This probably is very difficult to graph. You can't isolate the variables. There is no way to get all of the x on one side and all of the ys on the other.