Let the numbers be
\[a, \; b, \; c\]
so that \[a+b+c=K\]
for some constant \[K\]
. We want to maximise \[P=abc\]
.Rearrange
\[a+b+c=K\]
to give \[a=K-b-b\]
then \[P=(K-b-c)bc=Kbc-b^2c-bc^2\]
.For a maximum the partial derivatives with respect to
\[b\]
and \[c\]
will be zero, so\[\frac{\partial P}{\partial b}=Kc-2bc-c^2=c(K-2b-c)=0\]
\[\frac{\partial P}{\partial c}=Kb-b^2-2bc=b(K-b-2c)0\]
Obviously
\[b, \; c \neq 0\]
since then \[P=0\]
so \[K-2b-c=K-b-2c=0\]
.The solution is
\[b=c= \frac{K}{3}\]
so then \[a= \frac{K}{3}\]
and the maximum value of \[P\]
is \[\frac{K}{3} \times \frac{K}{3} \times \frac{K}{3} = \frac{K^3}{27}\]
.We can use the same method for any number of variables.