Find all points on the portion of the plane x+y+z=5 in the first octant at which f(x, y, z) = xy2z2 has a maximum value.

Lagrange multipliers:

[tex]L(x,y,z,\lambda)=xy^2z^2+\lambda(x+y+z-5)[/tex]

[tex]L_x=y^2z^2+\lambda=0[/tex]
[tex]L_y=2xyz^2+\lambda=0[/tex]
[tex]L_z=2xy^2z+\lambda=0[/tex]
[tex]L_\lambda=x+y+z-5=0[/tex]

[tex]\lambda=-y^2z^2=-2xyz^2=-2xy^2z[/tex]

[tex]-y^2z^2=-2xyz^2\implies y=2x[/tex] (if [tex]y,z\neq0[/tex])

[tex]-y^2z^2=-2xy^2z\implies z=2x[/tex] (if [tex]y,z\neq0[/tex])

[tex]-2xyz^2=-2xy^2z\implies z=y[/tex] (if [tex]x,y,z\neq0[/tex])

In the first octant, we assume [tex]x,y,z>0[/tex], so we can ignore the caveats above. Now,

[tex]x+y+z=5\iff x+2x+2x=5x=5\implies x=1\implies y=z=2[/tex]

so that the only critical point in the region of interest is (1, 2, 2), for which we get a maximum value of [tex]f(1,2,2)=16[/tex].

We also need to check the boundary of the region, i.e. the intersection of [tex]x+y+z=5[/tex] with the three coordinate axes. But in each case, we would end up setting at least one of the variables to 0, which would force [tex]f(x,y,z)=0[/tex], so the point we found is the only extremum.