1. **State the problem:** We want to design an open rectangular box with a square base of side length $b$ cm and height $h$ cm such that the volume is fixed at 256 cm³. We need to find the dimensions that minimize the surface area. 2. **Write down the formulas:** - Volume $V = b^2h = 256$ - Surface area $S$ (open box) = area of base + area of 4 sides = $b^2 + 4bh$ 3. **Express $h$ in terms of $b$ using the volume constraint:** $$h = \frac{256}{b^2}$$ 4. **Substitute $h$ into the surface area formula:** $$S = b^2 + 4b \times \frac{256}{b^2} = b^2 + \frac{1024}{b}$$ 5. **Minimize $S$ by finding critical points:** Take derivative with respect to $b$: $$\frac{dS}{db} = 2b - \frac{1024}{b^2}$$ Set derivative to zero: $$2b - \frac{1024}{b^2} = 0$$ Multiply both sides by $b^2$: $$2b^3 = 1024$$ $$b^3 = 512$$ $$b = \sqrt[3]{512} = 8$$ 6. **Find $h$ using $b=8$:** $$h = \frac{256}{8^2} = \frac{256}{64} = 4$$ 7. **Verify minimum by second derivative test:** $$\frac{d^2S}{db^2} = 2 + \frac{2048}{b^3}$$ At $b=8$: $$2 + \frac{2048}{512} = 2 + 4 = 6 > 0$$ So $S$ has a minimum at $b=8$. **Final answer:** The dimensions that minimize surface area are base side length $b=8$ cm and height $h=4$ cm.