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

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

Printable View

- Dec 30th 2010, 12:36 AMSambitconstraint maximization
Given $\displaystyle x+y=2$. Then what will be the maximum value of $\displaystyle x^3y^3(x^3+y^3)$?

Intuitively, I think it will be maximum when $\displaystyle x=y=1$. But can not put it mathematically. Any help? - Dec 30th 2010, 01:06 AMMoo
Hello,

Do you know Lagrange's multiplier ? - Dec 30th 2010, 01:54 AMHallsofIvy
One method is to use the first equation to replace y with 2- x. Then the function to be minimized becomes $\displaystyle x^3(2- x)^3(x^2+ (2- x)^3)$. 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 $\displaystyle <\lambda, \lambda>$ so, here, that just says that the two components of $\displaystyle grad(x^3y^3(x^3+ y^3)$ are equal:

$\displaystyle 3x^2y^3(x^2+ y^3)+ 3x^5y^3= 3x^3y^2(x^3+ y^3)+ 3x^3y^5$. 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 $\displaystyle x^2y^2$. - Dec 30th 2010, 03:03 AMCaptainBlack
- Dec 30th 2010, 06:43 AMSambit
As far as I know, Lagrange's fucntion is $\displaystyle x^3y^3(x^3+y^3)+\lambda(x+y-2)$ where $\displaystyle \lambda$ is Lagrange's multiplier. Differentiating this function partially wrt $\displaystyle x$ and $\displaystyle y$ equating them with $\displaystyle 0$, we get $\displaystyle 3x^2y^3(3x^2)+\lambda=0$ and $\displaystyle 3x^3y^2(3y^2)+\lambda=0$. Solving, we have $\displaystyle x=y=1$. Is this correct?