if x+y=1 then what is the min. value of x^3+y^3+xy ?

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- Feb 23rd 2013, 04:26 AM #1

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- Feb 23rd 2013, 05:14 AM #2

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- Feb 24th 2013, 02:47 AM #3

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- Feb 24th 2013, 03:35 AM #4

- Feb 24th 2013, 06:27 AM #5

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## Re: Minimium value

Another way to do it: $\displaystyle \nabla x+ y= <1, 1>$, $\displaystyle \nabla x^3+ y^3+ xy= <3x^2+ y, 3y^2+ x>$.

We must have $\displaystyle 3x^2+ y= \lambda$, $\displaystyle 3y^2+ x= \lambda$ for some number $\displaystyle \lambda$, as well as x+y= 1.

So $\displaystyle 3x^2+ y= 3y^2+ x$ which says that $\displaystyle 3x^2- 3y^2+ y- x= 3(x- y)(x+ y)- (x- y)= 0$.

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

If $\displaystyle x\ne y$ then $\displaystyle x- y\ne 0$ 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.

- Feb 24th 2013, 07:04 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 $\displaystyle x^3+ (1- x)^3+ x(1- x)= x^3+ 1- 3x+ 3x^2- x^3+ x- x^2= 2x^2- 2x+ 1$. We can write that as $\displaystyle 2(x^2- x)+ 1 $ and, completing the square, $\displaystyle 2(x^2- x+ 1/4-1/4)+ 1= 2(x^2- x+ 1/4)- 1/2+ 1= 2(x- 1/2)^2+ 1/2$. Since a square is never negative, that will have its minimum value $\displaystyle 2(1/2- 1/2)+ 1/2= 2(0)+ 1/2= 1/2$ when x= 1/2.