constraint maximization

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December 30th 2010, 12:36 AM
Sambit

constraint maximization

Given $x+y=2$ . Then what will be the maximum value of $x^3y^3(x^3+y^3)$ ?

Intuitively, I think it will be maximum when $x=y=1$ . But can not put it mathematically. Any help?
December 30th 2010, 01:06 AM
Moo

Hello,

Do you know Lagrange's multiplier ?
December 30th 2010, 01:54 AM
HallsofIvy

One method is to use the first equation to replace y with 2- x. Then the function to be minimized becomes $x^3(2- x)^3(x^2+ (2- x)^3)$ . You could either multiply that out or use the product rule to differentiate it.

If you are used to working with two or more variables and know "Lagrange's multiplier method", you can say that grad(x+ y)= <1, 1> while [itex]grad(x^3y^3(x^3+ y^3))= <3x^2y^3(x^3+ y^3+ 3x^5y^3, 3x^3y^2(x^3+ y^3)+ 3x^3y^5>[/tex]. "Lagrange's multiplier method" says that the max or min of one function subject to the constraint that the other be a constant occurs where one of those gradients is a multiple of the other. Since grad(x+ y)= <1, 1>, any multiple is of the form $<\lambda, \lambda>$ so, here, that just says that the two components of $grad(x^3y^3(x^3+ y^3)$ are equal:
$3x^2y^3(x^2+ y^3)+ 3x^5y^3= 3x^3y^2(x^3+ y^3)+ 3x^3y^5$ . That, together with the constraint, x+ y= 2, gives you two equations to solve for x and y.

I would start by saying that, if neither x nor y is 0, I can divide through by $x^2y^2$ .
December 30th 2010, 03:03 AM
CaptainBlack

Quote:

Originally Posted by Sambit

Given $x+y=2$ . Then what will be the maximum value of $x^3y^3(x^3+y^3)$ ?

Intuitively, I think it will be maximum when $x=y=1$ . But can not put it mathematically. Any help?

If you do not know about Lagrange multipliers as Moo suggests you can substitute x=2-y from the constraint into the objective differentiate the result and set to zero then solve for x. This will give you the constrained local stationary points ..

CB
December 30th 2010, 06:43 AM
Sambit

As far as I know, Lagrange's fucntion is $x^3y^3(x^3+y^3)+\lambda(x+y-2)$ where $\lambda$ is Lagrange's multiplier. Differentiating this function partially wrt $x$ and $y$ equating them with $0$ , we get $3x^2y^3(3x^2)+\lambda=0$ and $3x^3y^2(3y^2)+\lambda=0$ . Solving, we have $x=y=1$ . Is this correct?