Answer
$f(x,y,z)=xyz^{2}$ has a maximum value subject to the constraint x+y+z=6; it is at the point $(\frac{3}{2},\frac{3}{2},3)$. The value of $(\frac{3}{2},\frac{3}{2},3)$ is $\frac{81}{4}$.
Work Step by Step
$f(x,y,z)=xyz^{2}$
The constraint becomes $x+y+z-6=0$ with $g(x,y,z)=x+y+z-6$
$F(x,y,z,\lambda)=f(x,y,z)-\lambda.g(x,y,z)$
$F(x,y,z,\lambda)=xyz^{2}-\lambda(x+y+z-6)$
$F(x,y,z,\lambda)=xyz^{2}-\lambda x -\lambda y -\lambda z + 6\lambda$
$F_{x}(x,y,z,\lambda)=yz^{2}-\lambda$
$F_{y}(x,y,z,\lambda)=xz^{2}-\lambda$
$F_{z}(x,y,z,\lambda)=2xyz-\lambda$
$F_{\lambda}(x,y,z,\lambda)=-x-y-z+6$
$(1) yz^{2}-\lambda =0 \rightarrow \lambda =yz^{2}$
$(2) xz^{2}-\lambda =0 \rightarrow \lambda =xz^{2}$
$(3) 2xyz-\lambda =0 \rightarrow \lambda =2xyz$
$(4) -x-y-z+6=0$
Set the expressions equal $yz^{2}=xz^{2}$
$x=y$
Set the expressions equal $2xyz=xz^{2}$
$z=2y$
Substitute $x=y, z=2y$ in Equation (4)
$-y-y-2y+6=0$
$y=\frac{3}{2} \rightarrow x=\frac{3}{2} \rightarrow z=3$
The value of f is $f(\frac{3}{2},\frac{3}{2},3)=\frac{81}{4}$
Let y=1 and x=1, z=4; then f(1,1,4)=16, which is smaller than $\frac{81}{4}$. Because a nearby point has a value smaller than $\frac{81}{4}$ , the value $\frac{81}{4}$ is probably not a minimum.
$f(x,y,z)=xyz^{2}$ has a maximum value subject to the constraint x+y+z=6; it is at the point $(\frac{3}{2},\frac{3}{2},3)$. The value of $(\frac{3}{2},\frac{3}{2},3)$ is $\frac{81}{4}$.
