## Calculus with Applications (10th Edition)

$f_{max}(30,30,30)=27000$
Let our three positive numbers be x, y and z. We know that $x + y + z = 90 \rightarrow g(x)=x+y+z-90$ Our objective function is $f(x,y,z)=xyz$ $F(x,y,z,λ)=f(x,y,z)-\lambda.g(x,y,z)=xyz-\lambda (x+y+z-90)=xyz -\lambda x -\lambda y -\lambda z +90\lambda$ $F_{x}(x,y,z,\lambda)=yz-\lambda$ $F_{y}(x,y,z,\lambda)=xz-\lambda$ $F_{z}(x,y,z,\lambda)=xy-\lambda$ $F_{\lambda}(x,y,z,\lambda)=-x-y-z+90$ $(1) yz-\lambda=0 \rightarrow \lambda =yz$ $(2) xz-\lambda =0 \rightarrow \lambda =xz$ $(3) xy-\lambda =0 \rightarrow \lambda =xy$ $(4) -x-y-z+90=0$ Set the expressions equal: $yz=xz$ $x=y$ Set the expressions equal: $xz=xy$ $z=y$ Substitute for x=y=z in Equation (4) $-y-y-y+90=0$ $y=30 \rightarrow x=30 \rightarrow z=30$ The value of f is $f(30,30,30)=27000$ Let x=29.9 and y=29.9, z=30.2 then $f(29.9,29.9,30.2)=\approx26999.102$. Because a nearby point has a value smaller than 27000, the value 27000 is probably not a minimum. Therefore $f_{max}(30,30,30)=27000$