## Calculus: Early Transcendentals 8th Edition

The dimensions of the rectangle with minimum perimeter are $x=y=10\sqrt 10$ m.
If the rectangle has dimensions $x$ and $y$, then its area is $xy=1000 m^{2}$, so $y=\frac{1000}{x}$. The perimeter function is $y=2x+2y = 2x+\frac{2000}{x}$, and we wish to minimize this. $P'(x)=2-\frac{2000}{x^{2}}$, so the only critical number is $x=10\sqrt 10$. To solve for $y$, we use $y=\frac{1000}{x}$, and find $y=10\sqrt 10$ as well, giving us the dimensions that minimize the perimeter.