Answer:
Step-by-step explanation:
let the length=x
width=y
P=2(x+y)=128
x+y=128/2=64
y=64-x
area A=xy=x(64-x)=64x-x²
[tex]\frac{dA}{dx} =64-2x\\\\\frac{dA}{dx} =0,gives~64-2x=0,x=32\\\frac{d^2A}{dx^2} =-2<0 ,at ~x=32[/tex]
so A or area is maximum if x=32
y=64-32=32
or it is a square of edge=32 meters.
