## Calculus with Applications (10th Edition)

$f(x,y,z)=xy+2xz+2yz$ 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.
$f(x,y,z)=xy+2xz+2yz$ 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.