Chapter 4 - Applications of the Derivative - 4.1 Maxima and Minima - 4.1 Exercises - Page 243: 54

$ 8 \times 8 $.

$P(x) = 2x+128x , x > 0$, so $P(x) = 2− 128x^2$ , which is zero when $x^2 = 64$, or when $x = 8$. So $(8, 32)$ is the only critical point. This does turn out to be a minimum, so the dimensions of the rectangle with minimal perimeter are $8 × 8$.