Find the dimensions of a rectangle with perimeter 1000 metres so that the area of the rectangle is a maximum.

Dear student

perimeter of rectangle = 2(length+breadth)=1000

let length=x, then breadth = 500-x

Area = x*(500-x)

$$\begin{gathered} A=x(500-x)=500x-x^2 \\ \frac{dA}{dx}=500-2x \\ \frac{dA}{dx}=0 or 500-2x=0 or x=250 \\ \frac{d^{2}A}{d{x}^2}=-2 \\ so area is maximum \\ area =250*(500-250)=62500 \end{gathered}$$

Regards