# CoV3

Find the maximum of $xy^{2}z^{2}\,$ subject to the constraint $x+y+z=12\,$.

Using Lagrange multipliers, make a new function:

$f=xy^{2}z^{2}+\lambda (x+y+z-12)\,$

Take the partials of $f\,$ with respect to $x\,$, $y\,$, and $z\,$ and set them equal to zero.

1) ${\frac {\partial f}{\partial x}}=y^{2}z^{2}+\lambda =0\,$

2) ${\frac {\partial f}{\partial y}}=2xyz^{2}+\lambda =0\,$

3) ${\frac {\partial f}{\partial z}}=2xy^{2}z+\lambda =0\,$

This is a trick. Notice that if you multiply the last three equations by $x\,$, $y\,$, and $z\,$ respectively and then add them, you can factor out a $(x+y+z)\,$ term that is equal to $12\,$. So multiplying,

$xy^{2}z^{2}+\lambda x=0\,$

$2xy^{2}z^{2}+\lambda y=0\,$

$2xy^{2}z^{2}+\lambda z=0\,$

Now add these equations to get

$5xy^{2}z^{2}+\lambda (x+y+z)=5xy^{2}z^{2}+12\lambda =0\,$. So,

$\lambda ={\frac {-5}{12}}xy^{2}z^{2}\,$

Plug this value of lambda back into 1), 2), and 3) to get:

$x={\frac {12}{5}},y={\frac {24}{5}},z={\frac {24}{5}}\,$