Area

Let ${a}$ and ${b}$ be sides of the fence such that : $a+b = 40$ . Now we can go on to use the HM- GM inequality to maximize ${ab}$ : $\frac{1}{2}( \frac{1}{a} + \frac{1}{b}) \geq\sqrt{\frac{1}{ab}} \Rightarrow\frac{1}{2}(\frac{a+b}{ab}) = \frac{10}{ab} \geq \frac{1}{\sqrt{ab}} \Rightarrow\ 20 \geq\sqrt{ab}$ $\therefore\ Max(ab) = 20^2 = 400$

For maximum area of known perimeter, the rectangle must be a square. Let $x$ and $y$ be the side lengths of the rectangle. The perimeter is $2x+2y=80$ or $x+y=40$ . From here, $x=y$ , so $2x=40$ or $x=20$ . The area therefore is $20(20)=400$ .

This is common sense: Among all the rectangles that have the same perimeter, the square has the largest surface.

Hence, the largest surface occurs when the rectangle is a square, whose edge is $\displaystyle\frac{80}{4} = 20$ metres. Its surface is: $20 \times 20 = \boxed{400} \; m^2$

Just an other approach. L and W are the sides of the rectangle. P =2(L + W)= 80 .. A = L * W = L *( 40 - L ).....dA/dL= 40 - 2L = 0.......L = 20 ,.........W =20. A = 400.
In all such problem, all sides equal would be the answer. Say with a six sided solid , for given surface area, a cube has the maximum volume. In an ellipsoid, with a given surface, a sphere has the greatest volume.

We can make a quadratic function to solve it. The perimeter of the rectangle (it can also be a square) of sides $x$ and $h$ is $2x+2h=80$ or $x+h=40$ .

$h=40-x$

The area of the rectangle is $A=xh$ . How $h=40-x$ we have $A=x(40-x)$ or $A=-{x}^{2}+40x$ . This quadratic function is a parabola with a maximum value because the coefficient $a$ is negative. Let's calculate the maximum value of this parabola (the parabola vertex):

$-\frac{b}{2a}=\frac{-40}{-1\cdot2}=\frac{40}{2}=20$

Substituting in the equation $A=-{x}^{2}+40x$ :

$A=-{20}^{2}+40\cdot20$ $A=-400+800$ $A=400$

Thus the maximum area that the farmer can obtain is 400m^2!