A given rectangular area is to be fenced off in a field whose length…

The correct answer is Option (2) → The length is twice its breadth $(L=2B)$.

Let A (positive constant) be the area of the given rectangular field, and $x( > 0)$ be the length of the field, then its breadth = $\frac{A}{x}$.

Let $l$ be the length of the fencing required, then

$l = x + 2.\frac{A}{x}$

Differentiating (i) w.r.t. x, we get

$\frac{dl}{dx}=1-\frac{2A}{x^2}$ and $\frac{d^2l}{dx^2}=\frac{4A}{x^3}$.

Now $\frac{dl}{dx}=0⇒1-\frac{2A}{x^2}=0$

$⇒x^2 = 2A$ but $x > 0 ⇒ x = \sqrt{2A}$.

Also $\left(\frac{d^2l}{dx^2}\right)_{x=\sqrt{2A}}=\frac{4A}{2A.\sqrt{2A}}>0$

$⇒l$ is minimum at $x=\sqrt{2A}$.

Therefore, fencing required is least when length of field = $\sqrt{2A}$

and breadth $=\frac{A}{\sqrt{2A}}=\frac{\sqrt{2A}}{2}=\frac{1}{2}$ of length of field.

Hence, the fencing is least when the length of the field is twice its breadth.