Answer:
The perimeter is minimum for Length and width both are [tex]\sqrt3[/tex].
Step-by-step explanation:
Area, A = 3 square metre
Let the length is L and width is W.
Area = L W
3 = L W.....(1)
The perimeter is given by
P = 2 (L + W)
Substitute the value of from (1)
[tex]P = 2 \left ( L +\frac{3}{L} \right )\\\\P = 2 L + \frac{6}{L}\\\\\frac{dP}{dL} = 2 - \frac{6}{L^2}\\\\Now\\\\\frac{dP}{dL}=0\\\\2 - \frac{6}{L^2} = 0\\\\L = \sqrt 3, W = \sqrt 3[/tex]
Now
[tex]\frac{d^2P}{dL^2}=\frac{12}{L^3}\\[/tex]
It is alays positive, so the perimeter is minimum.
