Rhombus and its diagonals

The area of the rhombus is $A = \frac{1}{2}d_1d_2$ , and if the sum of the $2$ diagonal lengths is $20$ , then $d_1 + d_2 = 20$ .

Substituting gives $A = \frac{1}{2}d_1(20 - d_1) = 10d_1 - \frac{1}{2}d_1^2$ .

The maximum area is when the derivative is equal to $0$ , or when $A' = 10 - d_1 = 0$ . Solving this gives $d_1 = 10$ .

Since $d_1 + d_2 = 20$ , solving $10 + d_2 = 20$ gives $d_2 = 10$ .

Finally, since $A = \frac{1}{2}d_1d_2$ , $A = \frac{1}{2}(10)(10) = \boxed{50}$ .

Let $a, b$ be the lengths of the diagonals of the rhombus. We know that $a + b = 20,$ and the area of a rhombus is given by $A = \dfrac{ab}{2}.$ Since $a, b$ are side lengths and are thus positive real numbers, we can apply the AM-GM inequality to get

$ab \leq \left ( \dfrac{a + b}{2} \right )^2 = \left ( \dfrac{20}{2} \right )^2 = 100.$

Therefore, the maximum possible area of the rhombus is $A \leq \dfrac{100}{2} = \boxed{50},$ which is achieved when $a = b = 10$ i.e. the rhombus is a square with diagonals of length 10.

Let $d_1$ and $d_2$ be the diagonals. So, the area is $\frac{1}{2}d_1d_2$ .

And by hypothesis, $d_1+d_2=20$ .

From Algebra, $4d_1d_2 = (d_1+d_2)^2-(d_1-d_2)^2$

$\implies \frac{1}{2}d_1d_2 = \frac{(d_1+d_2)^2-(d_1-d_2)^2}{8}= \frac{20^2-(d_1-d_2)^2}{8} = \frac{400-(d_1-d_2)^2}{8}$

So, the maximum value of $\frac{1}{2}d_1d_2$ occurs when $d_1-d_2=0$ , or equivalently, $d_1=d_2$ .

Hence the maximum value of area $= \frac{400}{8} = \boxed{50}$ .