Arectangle has a perimeter of 32 in. find the length and width of the rectangle under which the area is the largest. follow the steps: (a) let the width to be x and the length to be y , then the quantity to be maximized is (expressed as a function of both x

Let width and length be x and y respectively.
Perimeter (32in) =2x+2y=> 16=x+y => y=16-x
Area, A = xy = x(16-x) = 16x-x^2
The function to maximize is area: A=16 x-x^2
For maximum area, the first derivative of A =0 => A'=16-2x =0
Solving for x: 16-2x=0 =>2x=16 => x=8 in
And therefore, y=16-8 = 8 in