Do you know Lagrange's multiplier ?
One method is to use the first equation to replace y with 2- x. Then the function to be minimized becomes . 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 so, here, that just says that the two components of are equal:
. 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 .