Calculus with Applications (10th Edition)

$f(x,y,z)=xyz^{2}$ 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 $\frac{81}{4}$.
$f(x,y,z)=xyz^{2}$ 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 $\frac{81}{4}$.