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Please help me with this math problem.

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Someone on ATOT, has to have a high level undeder standing of math. Where are the math, physics and engineering Ph.D crew.
 
Originally posted by: 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.

Make sure you didn't miss anything else before asking ATOT to do your homework.
 
Originally posted by: StatsManD
Someone on ATOT, has to have a high level undeder standing of math. Where are the math, physics and engineering Ph.D crew.

Probably hanging out in Highly Technical.

I wouldn't bother them if I were you, though.
 
Originally posted by: chuckywang
Originally posted by: StatsManD
Someone on ATOT, has to have a high level undeder standing of math. Where are the math, physics and engineering Ph.D crew.

The case when c=0 is trivial.



Now prove as c goes to 1 the shaded area decreases monotonically. then prove this statement still holds true when we interchange x and y.
 
Originally posted by: StatsManD
Originally posted by: chuckywang
Originally posted by: StatsManD
Someone on ATOT, has to have a high level undeder standing of math. Where are the math, physics and engineering Ph.D crew.

The case when c=0 is trivial.



Now prove as c goes to 1 the shaded area decreases monotonically. then prove this statement still holds true when we interchange x and y.

That's pretty easy too. Assume the area doesn't decrease monotonically as c increases. Find a contradiction.
 
Originally posted by: StatsManD
Originally posted by: chuckywang
Originally posted by: StatsManD
Someone on ATOT, has to have a high level undeder standing of math. Where are the math, physics and engineering Ph.D crew.

The case when c=0 is trivial.



Now prove as c goes to 1 the shaded area decreases monotonically. then prove this statement still holds true when we interchange x and y.

Why woudln't it be true? there's nothing to prove. Everything condition you have on x and y (namely, only one, which is x+y<=1) is symmetrical wrt x and y.
 
Originally posted by: chuckywang
Originally posted by: StatsManD
Originally posted by: chuckywang
Originally posted by: StatsManD
Someone on ATOT, has to have a high level undeder standing of math. Where are the math, physics and engineering Ph.D crew.

The case when c=0 is trivial.



Now prove as c goes to 1 the shaded area decreases monotonically. then prove this statement still holds true when we interchange x and y.

That's pretty easy too. Assume the area doesn't decrease monotonically as c increases. Find a contradiction.

How would I find a contradiction.
 
Think about it this way. Let A1 be the area associated with c=c1. Let A2 be the area associated with c=c2. Assume c1 > c2.

Every (x,y) point in area A1 is also in area A2 (why? think about the inequality). Therefore, A2 >= A1.

Forget about the contradiction stuff I said. This is much more straightforward.
 
Use the Riemann-Cauchy Theorem and then integral the function. After that, use induction to prove it, using that equation after integration.
 
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