maximum area

Let the side lengths of the rectangle be $a$ and $b$ and its area be $A$ . Therefore, $a^2+b^2 = 20^2$ and $A=ab$ .

By AM-GM inequality, we have:

$\dfrac {a^2+b^2}{2} \ge ab \quad \Rightarrow A \le \dfrac {a^2+b^2}{2} = \dfrac {20^2}{2} = \boxed{200}$

The diagonal length is 20.

The maximum area of a rectangle per diagonal length is a square.

From the 45-45-90 special triangles we can derive that the ration of the side length of a square to its diagonal is $\frac{1}{\sqrt{2}}$ .

So the side length is $\frac{20}{\sqrt{2}}$ .

So the area is $\left(\frac{20}{\sqrt{2}}\right)^2$

So the area is 200.

Let x be the breadth of the Rectangle

The area of a rectangle is given by length×breadth.

The diagonal will be given by the expression

\sqrt{ {length}^{2} + {breadth}^{2} }

Given that diagonal is 20 units long.

We can write

\sqrt{ {length}^{2} + {breadth}^{2} } = 20

Length = \sqrt{400 - {breadth}^{2} }

Area=length×breadth

Area = breadth \times \sqrt{400 - {breadth}^{2} }

Then we need to maximize the function:

f(x) = x*sqrt(400-x^2) using Calculus.

2ab = a^2+b^2 - (a-b)^2 = 400 - (a-b)^2. Therefore, max. value of ab is 200 when a=b.