Acquired Area

Let the side lengths of the rectangle be $a$ and $b$ . Then the perimeter is $2a+2b = 200$ , $\implies a+100$ . By AM-GM inequality we have:

$\begin{aligned} a + b & \ge 2\sqrt{ab} \\ 100 & \ge 2 \sqrt{ab} \\ ab & \le \left(\frac {100}2 \right)^2 = 2500 \end{aligned}$

Since $ab$ is the area of the rectangle. The maximum area is $\boxed{2500}$ when $a=b = 50$ .

With a fixed perimeter given, the area is largest when all the sides are equal i.e they form a square. Since the given perimeter is 200 and there are 4 sides, each individual side of the square is 50. Hence the area of this square is

$50 \times 50 = \boxed{2500}$

$\text{We could solve this problem using Calculus:}$

$\text{Let}$ $x$ $\text{and}$ $y$ $\text{denote the lengths of the rectangle,}$ $P$ $\text{the perimeter, and}$ $A$ $\text{the area}$

$\text{We know}$ $P=200$ $\Longrightarrow$ $2x+2y=200$ $\Longrightarrow$ $x+y=100$ $\Longrightarrow$ $y=100-x$

$\text{We are interested on maximize}$ $A=xy$ $\text{. Let´s substitute}$ $y$ $\text{on this equation}$ $\text{. We obtain:}$

$A=x(100-x)$ $\Longrightarrow$ $A=100x-x^2$

$\text{Derivating and equating to 0 we can search our candidates for maximize A:}$

$A' = 100 - 2x = 0$ $\Longrightarrow$ $\boxed{x=50}$

$\text{Using the second derivative, we can prove:}$ $A'' = -2 < 0$ $\Longrightarrow$ $x=50$ $\text{maximize A}$

$\text{We can calculate}$ $y:$

$y=100-x$ $\Longrightarrow$ $\boxed{y=50}$

$\text{Finally, we can say that the maximum area should be: }$

$A=xy$ $\Longrightarrow$ $A=50 * 50$ $\Longrightarrow$ $\boxed{A=2500}$