Min =1/2 when x =1/2

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- February 23rd 2013, 05:26 AM #1

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- February 23rd 2013, 06:14 AM #2

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- February 24th 2013, 03:47 AM #3

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- February 24th 2013, 04:35 AM #4

- February 24th 2013, 07:27 AM #5

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## Re: Minimium value

Another way to do it: , .

We must have , for some number , as well as x+y= 1.

So which says that .

If x= y, from x+y= 1 we have x= y= 1/2.

If then so we can divide by it to get 3(x+ y)- 1= 0 which says x+y= 1/3 contradicting x+y= 1.

Thus x= y= 1/2 is the only solution.

- February 24th 2013, 08:04 AM #6

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## Re: Minimium value

Yet

**another**way, and probably simplest:

From x+ y= 1, y= 1- x. Putting that into [tex]x^3+ y^3+ xy[/itex] as Prove It suggested, we have . We can write that as and, completing the square, . Since a square is never negative, that will have its minimum value when x= 1/2.