Answer
$x=10\sqrt{10}$ and $y= 10\sqrt{10}$
Work Step by Step
Let $x$ be the length and width be $y $ then the area given by
$$A=xy\ \ \Rightarrow \ 1000=xy \ \Rightarrow \ y=\frac{1000}{x}$$
Since perimeter give by
$$P=2x+2y=2x+\frac{2000}{x}$$
The goal is to minimize $P$ , since
$$ P'(x)=2-\frac{2000}{x^2}$$
Then $P'(x)=0$ for $x=10\sqrt{10}$ , since $P'(x)<0$ for $x<10\sqrt{10}$ , $P'(x)>0$ for $x>10\sqrt{10}$, hence $P(x)$ has minimum at $x=10\sqrt{10}$ and $y= 10\sqrt{10}$
