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**John92****Guest**

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.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,727

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.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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**John92****Guest**

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****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,727

If x,y were in the interval of 0 and 1?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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**John92****Guest**

That's right

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,727

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).

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,015

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.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,727

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.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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