Answer
$f(x,y,z)=xy+2xz+2yz$ has a maximum value subject to the constraint xyz=32; it is at the point $(4,4,2)$. The value of $(4,4,2)$ is 48.
Work Step by Step
$f(x,y,z)=xy+2xz+2yz$
The constraint becomes $xyz-32=0$ with $g(x,y,z)=xyz-32$
$F(x,y,z,\lambda)=f(x,y,z)-\lambda.g(x,y,z)$
$=xy+2xz+2yz-\lambda(xyz-32)=xy+2xz+2yz-\lambda xyz + 32\lambda$
$F_{x}(x,y,z,\lambda)=y+2z-\lambda yz$
$F_{y}(x,y,z,\lambda)=x + 2z-\lambda xz$
$F_{z}(x,y,z,\lambda)=2x + 2y-\lambda xy$
$F_{\lambda}(x,y,z,\lambda)=-xyz+32$
$(1) y+2z-\lambda yz =0 \rightarrow \lambda =\frac{y+2z}{yz}$
$(2) x + 2z-\lambda xz =0 \rightarrow \lambda =\frac{x+2z}{xz}$
$(3) 2x + 2y-\lambda xy =0 \rightarrow \lambda =\frac{2x+2y}{xy}$
$(4) -xyz+32=0$
Set the expressions equal $\frac{y+2z}{yz}=\frac{x+2z}{xz}$
$xz(y+2z)=yz(x+2z)$
$xyz+2xz^{2}=xyz+2yz^{2}$
$x=y$
Set the expressions equal $\frac{x+2z}{xz}=\frac{2x+2y}{xy}$
$xy(x+2z)=xz(2x+2y)$
$x^{2}y+2xyz=2x^{2}z+2xyz$
$y=2z \rightarrow z=\frac{1}{2}y$
Substitute for $x=y, z=\frac{1}{2}y$ in Equation (4)
$-y.y.\frac{1}{2}y+32=0$
$-\frac{1}{2}y^{3}=-32$
$y=4 \rightarrow x=4 \rightarrow z=2$
The value of f is $f(4,4,2)=48$
Let y=3.9 and x=3.9, $z=2.1$ then $f(3.9,3.9,2.1)=47.97$ which is smaller than 48. Because a nearby point has a value smaller than 48, the value 48 is probably not a minimum.
$f(x,y,z)=xy+2xz+2yz$ has a maximum value subject to the constraint xyz=32; it is at the point $(4,4,2)$. The value of $(4,4,2)$ is 48.