Answer
With $x=8, y=16, f(x,y) =3xy^{2}$ is maximized at 6144
Work Step by Step
We are given $f(x,y)=3xy^{2}$
with $x+y=24$ so $y=24-x$
$f(x,y)=3x(24-x)^{2}=3x(576-48x+x^{2})=1728x-144x^{2}+3x^{3}$
Now differentiate f(x) with respect to x and equate it to '0' as shown below
$f_{x}(x,y)=1728-288x+9x^{2} = 0$
$x=24 \rightarrow y=0 \rightarrow $ The value of $3xy^{2}$ is 0
or $x=8 \rightarrow y=16 \rightarrow $ The value of $3xy^{2}$ is 6144
Since $6144\gt0$, with $x=8, y=16, f(x,y) =3xy^{2}$ is maximized at 6144
