Minimium value

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February 23rd 2013, 04:26 AM
RuyHayabusa

Minimium value

if x+y=1 then what is the min. value of x^3+y^3+xy ?
February 23rd 2013, 05:14 AM
MINOANMAN

Re: Minimium value

Min =1/2 when x =1/2
February 24th 2013, 02:47 AM
RuyHayabusa

Re: Minimium value

Step-by-step solution please.
February 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.
February 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.
February 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.

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