Show that the rectangle of largest possible area, for a given perimeter, is a square.
Show that the rectangle of largest possible area, for a given perimeter, is a square.
Let X be the length and y be the breadth of rectangle and P be it's perimeter.
P=2(x+y)
P/2=x+y
y=P/2-x
If A be it's area then,
A=x.y
A=x.(P/2-x)
A=Px/2 - x2
dA/dx=P/2 - 2x
d2A/dx2= -2<0 [Maximum]
Since
dA/dx=0
P/2 - 2x=0
x=P/4
Also, y=P/2 - x =P/2-P/4 = P/4
Since length and breadth are equal, the rectangle is a square.