Question

show that among the rectangles of given area the square has the least perimeter

Answer 1

Let A be the area of rectangle
A=xy
y=A/x
x&y be the sides of rectangle
P=2(x+A/x)
dP/dx=2(1-A/x²)
put dP/dx=0
2(1-A/x²)=0
1-A/x²=0
1=A/x²
or A=x²
A=xy
so,xy=x²
y=x²/x=x
y=x


d²P/dx²=2(-(-2A)/x³)
=4A/x³
=4x²/x³=4/x
d²P/dx² is positive when A=x²or y=x
perimeter is least when x=y
so,when A=x² or both sides are equal
it will be a square
i.e.square has the least perimeter