Question

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

Answer 1

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