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

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

let length=x, then breadth = 500-x

Area = x*(500-x)

$A=x(500-x)=500x-{x}^{2}\phantom{\rule{0ex}{0ex}}\frac{dA}{dx}=500-2x\phantom{\rule{0ex}{0ex}}\frac{dA}{dx}=0or500-2x=0orx=250\phantom{\rule{0ex}{0ex}}\frac{{d}^{2}A}{d{x}^{2}}=-2\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}soareaismaximum\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}area=250*(500-250)=62500$

Regards

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