Given that the perimeter of a rectangle is 50, find the largest possible area of this rectangle.

The answer is 156.25.

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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 $\boxed{156.25}$ .

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 $A=156.25$