Answer
$f_{max}(30,30,30)=27000$
Work Step by Step
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 $f_{max}(30,30,30)=27000$
