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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}}\,

Calculus of Variations

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