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

Results 1 to 6 of 6

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

- Joined
- Feb 2013
- From
- US
- Posts
- 37
- Thanks
- 1

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

- Joined
- Feb 2013
- From
- Saudi Arabia
- Posts
- 440
- Thanks
- 86

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

- Joined
- Feb 2013
- From
- US
- Posts
- 37
- Thanks
- 1

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

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

- Joined
- Apr 2005
- Posts
- 15,536
- Thanks
- 1391

## 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, 07:04 AM #6

- Joined
- Apr 2005
- Posts
- 15,536
- Thanks
- 1391

## 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.