we are given
[tex]A=24x-x^2[/tex]
where
A is area of rectangle
x is width of rectangle
Since, we have to maximize area
so, we will find derivative
[tex]A'=24\times 1-2x[/tex]
[tex]A'=24-2x[/tex]
now, we can set it to 0
and then we can solve for x
[tex]A'=24-2x=0[/tex]
[tex]24-2x=0[/tex]
[tex]x=12[/tex]
The width which gives you the maximum area is 12 feet........Answer
now, we can plug x=12 to find area
[tex]A=24(12)-(12)^2[/tex]
[tex]A=288-144[/tex]
[tex]A=144ft^2[/tex]
So, the maximum area is
[tex]A=144ft^2[/tex]............Answer