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- bobbym
- 2012-11-11 10:49:13
Hi;
that at least one of the values, A, B, C would be greater than D at every value of x and y
Looks like he is saying A or B or C is always greater in that interval.
- anonimnystefy
- 2012-11-11 10:04:55
I don't think the point is to prove that one of those is always greater than D, but rather at least one of those will be greater than D for some x and y.
- bobbym
- 2012-11-11 09:53:56
Hi;
A is out, it is not always greater than D in the closed interval (0,1).
B is out, it is not always greater than D in the closed interval (0,1).
C is out, it is not always greater than D in the closed interval (0,1).
- John92
- 2012-11-11 09:47:20
That's right
- bobbym
- 2012-11-11 09:39:30
If x,y were in the interval of 0 and 1?
- John92
- 2012-11-11 09:21:42
Sorry, maybe ignore that bit, i'm not sure that even makes sense anyway - i was just speculating about how i'd demonstrate that at least one of the values, A, B, C would be greater than D at every value of x and y.
- bobbym
- 2012-11-11 09:16:16
Hi;
D surface is always below at least one other
I am not following you here. Please post the exact question. Wording is very important in mathematics.
- John92
- 2012-11-11 09:09:22
Hey guys, i'm doing an econ assignment but i'm either going about the question in completely the wrong way - or im stuck on some math that i'm unable to do. Assuming it's the second (because i dont just want to post the question and get you to do the whole thing for me), here's what im stuck on.
A = 9xy
B = 9x + 9y - 18xy
C = 9 - 9x - 9y + 9xy
D = 12xy - 6x - 6y + 6
I want to prove that there are no values of x and y (where x and y are both values between 0 and 1) for which D is the greatest.
Is that possible to do numerically? Only thing i can think of is having some software plot a 3d graph and show that the D surface is always below at least one other - obviously thats no good for helping me do my homework though lol.