# Minimium value

• Feb 23rd 2013, 04:26 AM
RuyHayabusa
Minimium value
if x+y=1 then what is the min. value of x^3+y^3+xy ?
• Feb 23rd 2013, 05:14 AM
MINOANMAN
Re: Minimium value
Min =1/2 when x =1/2
• Feb 24th 2013, 02:47 AM
RuyHayabusa
Re: Minimium value
• Feb 24th 2013, 03:35 AM
Prove It
Re: Minimium value
Quote:

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

y = 1 - x

Substitute this into the second equation, then evaluate where the derivative of the second function is 0.
• Feb 24th 2013, 06:27 AM
HallsofIvy
Re: Minimium value
Another way to do it: $\nabla x+ y= <1, 1>$, $\nabla x^3+ y^3+ xy= <3x^2+ y, 3y^2+ x>$.

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

So $3x^2+ y= 3y^2+ x$ which says that $3x^2- 3y^2+ y- x= 3(x- y)(x+ y)- (x- y)= 0$.
If x= y, from x+y= 1 we have x= y= 1/2.

If $x\ne y$ then $x- y\ne 0$ 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
HallsofIvy
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 $x^3+ (1- x)^3+ x(1- x)= x^3+ 1- 3x+ 3x^2- x^3+ x- x^2= 2x^2- 2x+ 1$. We can write that as $2(x^2- x)+ 1$ and, completing the square, $2(x^2- x+ 1/4-1/4)+ 1= 2(x^2- x+ 1/4)- 1/2+ 1= 2(x- 1/2)^2+ 1/2$. Since a square is never negative, that will have its minimum value $2(1/2- 1/2)+ 1/2= 2(0)+ 1/2= 1/2$ when x= 1/2.