The maximum area of a rectangle (in $m^{2}$ ) having perimeter $1000 m$ is

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Solution

Let the length of the rectangle be $x m$ , width be $y m$ and area be $A m^{2}$
The perimeter of the rectangle is $1000 m$ ,
$2 x + 2 y = 1000$
$\begin{matrix} \Rightarrow y = 500 - x \\ (1) \end{matrix}$
Area of rectangle, $A = x y$
Thus $\begin{matrix} A = x (500 - x) = 500 x - x^{2} \\ (2) \end{matrix}$
To get the stationary point at which $A$ will be maximum, we will differentiate equation $(2)$ ,
$\frac{d A}{d x} = 500 - 2 x = 0$
$x = 250 m$
Substituting the value of $x$ in $(1)$ , we get
$y = 250 m$
Hence maximum area, $A = x y = 250 \times 250 = 62500 m^{2}$

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