Let's assume total cost = \(200L+100B+100L+300L+200B……(1)\)
Area of rectangle = \(L \times B=600 …..(2)\)
We know that, \(AM> GM\)
\(=300L+200B>2 \times60000\)
Total cost is \(> 120000\)
∴ The lowest possible cost of fencing all four sides in INR is 120000.
Let the length and breadth be \(x\) and \(y\) respectively.
Then the area of the region,
\(xy = 60000\) \(……… (i)\)
Then the cost of fencing,
\(200x + 100x + 100y + 100y\)
\(⇒300x + 200y\)
We know that,
Arithmetic mean \(≥\) Geometric mean
\(⇒\frac {300x+200y}{2}≥\sqrt {300x×200y}\)
\(⇒300x+200y≥2\sqrt {300x×200y}\)
\(⇒300x+200y≥2\sqrt {300×200 ×\ xy}\)
\(⇒300x+200y≥2\sqrt {300×200 ×\ 6000}\) [From eq \((i)\)]
\(⇒300x+200y≥2\sqrt {3600000000}\)
\(⇒300x + 200y ≥ 2 \times 60000\)
\(⇒300x + 200y ≥ 120000\)
Thus, we can say the cost will always be greater than \(120000\).
So, the correct option is (C): \(120000\).