Answer
$f_{max}(80,80,80)=512000$
Work Step by Step
Let our three positive numbers be x, y and z. We know that
$x + y + z =240 \rightarrow g(x)=x+y+z-240$
Our objective function is $f(x,y,z)=xyz$
$F(x,y,z,λ)=f(x,y,z)-\lambda.g(x,y,z)=xyz-\lambda (x+y+z-240)=xyz -\lambda x -\lambda y -\lambda z +240\lambda$
$F_{x}(x,y,z,\lambda)=yz-\lambda$
$F_{y}(x,y,z,\lambda)=xz-\lambda$
$F_{z}(x,y,z,\lambda)=xy-\lambda$
$F_{\lambda}(x,y,z,\lambda)=-x-y-z+240$
$(1) yz-\lambda=0 \rightarrow \lambda =yz$
$(2) xz-\lambda =0 \rightarrow \lambda =xz$
$(3) xy-\lambda =0 \rightarrow \lambda =xy$
$(4) -x-y-z+240=0$
Set the expressions equal:
$yz=xz$
$x=y$
Set the expressions equal:
$xz=xy$
$z=y$
Substitute for x=y=z in Equation (4)
$-y-y-y+240=0$
$y=80 \rightarrow x=80 \rightarrow z=80$
The value of f is $f(80,80,80)=512000$
Let x=79.9 and y=79.9, z=80.2 then $f(79.9,79.9,80.2)=\approx 511997.602$. Because a nearby point has a value smaller than 512000, the value 512000 is probably not a minimum.
Therefore $f_{max}(80,80,80)=512000$
