A área do retângulo. In English!

The quadrilateral with the greatest area for a given perimeter is a square. From this we can say that each side has a length of $\frac{50}{4}$ To find area, solve $\frac{50}{4}*\frac{50}{4}$ which equals $\boxed{156.25}$ .

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EXTRA PROOF

If you don't believe me that a square will give the greatest area, you can use calculus to prove it

PERIMETER

$P=2a+2b=50$ , where $a$ and $b$ are the side lengths.

AREA

$A=a*b$

$2a+2b=50 \Rightarrow a=25-b$ [rearrange and simplify]

$A=(25-b)b$ [substitute into area formula]

$A=25b-b^2$ [expand]

$A'=25-2b$ [differentiate]

Let $A=0$ (to find minimum)

$0=25-2b$

$2b=25$ [rearrange]

$b=12.5$ [solve for $b$ ]

$2a+25=50$ [substitute $b$ into perimeter equation]

$2a=25$ [rearrange]

$a=12.5$ [solve for $a$ ]

$a=b$ therefore shape is a square and $A=156.25$

Using Lagrangian method, L= a b - m(2a + 2b -50) dL/da = b - 2m = 0 max since dL/da = -2 < 0 dl/db = a - 2m = 0 solving gives a=b, thus 4a = 50 a = 12.5 Area is a^2 = 12.5^2 = 156.25 m is the lagrange multiplier