A farmer has a 200 m fencingmaterial to enclose two adjacent sides of a rectangular...

Let area be $A$ and the sides of rectangular field be $x$ and $y$;
So,

$A=x\cdot y$

Now, one side of the rectangle is already made with a fence.
There are $4$ sides, two sides of $x$ meters and two sides of $y$ meters. Let a side of $y$ meters be already fenced.

Then, the remaining three sides are to be fenced with a fence of $200 \mathrm{m}$ length,

So,

$2x+y=200$

$y=200-2x$
So,

$A=x(200-2x)$

$A=200x-2x^2$

For area to be maximum,

$\frac{dA}{dx}=0$ and $\frac{d^{2}A}{{dx}^2}<0$

Now,

$\frac{dA}{dx}=200-4x$

$200-4x=0$

$x=50$

Now,

$\frac{d^{2}A}{{dx}^2}=-4$

Which is negative

Thus, $x=50$ is a maximum.

Now,

$2x+y=200$

$2\left(50\right)+y=200$

$y=100$

Thus, the required area is:

$A=x\cdot y=50\cdot 100=5000 {\mathrm{m}}^2$