The length of a rectangle is three times its width. Given that its perimeter and area are both numerically equal to $k>0$, find $k$.
Let $a$ be the width of the rectangle. Then the length of the rectangle is $3 a$, so the perimeter is $2(a+3 a)=8 a$, and the area is $3 a^{2}$. Since the length is numerically equal to the width, we know that \[ 8 a=3 a^{2}=k \] Because $k>0$, the rectangle is non-degenerate. It follows that $8=3 a$, so $a=\frac{8}{3}$. Therefore, $k=\frac{64}{3}$. $\fbox{\frac{64}{3}}$.