Answer
With $x=32, y=16, f(x,y) =5x^{2}y+10$ is maximized at 81930
Work Step by Step
We are given $f(x,y)=5x^{2}y+10$
with $x+y=48$ so $y=48-x$
$f(x,y)=5x^{2}(48-x)+10=240x^{2}-5x^{3}+10$
Now differentiate f(x) with respect to x and equate to '0' as shown below
$f_{x}(x,y)=480x-15x^{2} = 0$
$x=32 \rightarrow y=16 \rightarrow $ The value of $5x^{2}y+10$ is 81930
or $x=0 \rightarrow y=48 \rightarrow $ The value of $5x^{2}y+10$ is 10
Since $81930\gt10$, with $x=32, y=16, f(x,y) =5x^{2}y+10$ is maximized at 81930
