1. **State the problem:** We need to design a rectangular cargo container with a fixed volume of 12 m³. The length $L$ must be 1.5 times the width $W$, and we want to minimize the total surface area $S$. 2. **Given:** - Volume $V = 12$ m³ - Length $L = 1.5W$ - Height $H$ unknown 3. **Formulas:** - Volume of a rectangular box: $$V = L \times W \times H$$ - Surface area of a rectangular box: $$S = 2(LW + LH + WH)$$ 4. **Express variables:** Since $L = 1.5W$, substitute into volume formula: $$12 = 1.5W \times W \times H = 1.5W^2H$$ Solve for $H$: $$H = \frac{12}{1.5W^2} = \frac{8}{W^2}$$ 5. **Express surface area $S$ in terms of $W$ only:** $$S = 2(LW + LH + WH) = 2(1.5W \times W + 1.5W \times H + W \times H)$$ Simplify inside parentheses: $$= 2(1.5W^2 + 1.5WH + WH) = 2(1.5W^2 + 2.5WH)$$ Substitute $H = \frac{8}{W^2}$: $$S = 2\left(1.5W^2 + 2.5W \times \frac{8}{W^2}\right) = 2\left(1.5W^2 + \frac{20}{W}\right) = 3W^2 + \frac{40}{W}$$ 6. **Minimize $S$ by finding critical points:** Take derivative with respect to $W$: $$\frac{dS}{dW} = 6W - \frac{40}{W^2}$$ Set derivative to zero: $$6W - \frac{40}{W^2} = 0 \implies 6W = \frac{40}{W^2} \implies 6W^3 = 40 \implies W^3 = \frac{40}{6} = \frac{20}{3}$$ 7. **Solve for $W$:** $$W = \sqrt[3]{\frac{20}{3}} \approx 1.873$$ meters 8. **Find $L$ and $H$:** $$L = 1.5W = 1.5 \times 1.873 = 2.81$$ meters $$H = \frac{8}{W^2} = \frac{8}{(1.873)^2} = \frac{8}{3.51} \approx 2.28$$ meters 9. **Verify volume:** $$V = LWH = 2.81 \times 1.873 \times 2.28 \approx 12$$ m³ (as required) 10. **Conclusion:** The most efficient container dimensions to minimize surface area with volume 12 m³ and length 1.5 times width are approximately: - Width $W \approx 1.87$ m - Length $L \approx 2.81$ m - Height $H \approx 2.28$ m These dimensions minimize material usage and weight while meeting volume and design constraints.