1. **Problem Statement:** We have a closed cuboid with volume 150 cm³. Height is $h$, width is $w$, and length is $3w$. We want to minimize the surface area. 2. **Express height $h$ in terms of $w$ using volume:** Volume formula: $$V = \text{length} \times \text{width} \times \text{height}$$ Given: $$150 = 3w \times w \times h = 3w^2 h$$ Solve for $h$: $$h = \frac{150}{3w^2} = \frac{50}{w^2}$$ 3. **Express surface area $A$ in terms of $w$:** Surface area of cuboid: $$A = 2(lw + lh + wh)$$ Substitute $l=3w$ and $h=\frac{50}{w^2}$: $$A = 2(3w \times w + 3w \times \frac{50}{w^2} + w \times \frac{50}{w^2})$$ Simplify each term: $$3w \times w = 3w^2$$ $$3w \times \frac{50}{w^2} = \frac{150w}{w^2} = \frac{150}{w}$$ $$w \times \frac{50}{w^2} = \frac{50w}{w^2} = \frac{50}{w}$$ So, $$A = 2\left(3w^2 + \frac{150}{w} + \frac{50}{w}\right) = 2\left(3w^2 + \frac{200}{w}\right) = 6w^2 + \frac{400}{w}$$ 4. **Minimize surface area $A(w)$:** Find derivative: $$A'(w) = 12w - \frac{400}{w^2}$$ Set derivative to zero for critical points: $$12w - \frac{400}{w^2} = 0$$ Multiply both sides by $w^2$: $$12w^3 - 400 = 0$$ $$12w^3 = 400$$ $$w^3 = \frac{400}{12} = \frac{100}{3}$$ $$w = \sqrt[3]{\frac{100}{3}} \approx 3.17$$ 5. **Find corresponding $h$:** $$h = \frac{50}{w^2} = \frac{50}{(3.17)^2} \approx \frac{50}{10.05} \approx 4.97$$ 6. **Calculate minimum surface area:** $$A = 6w^2 + \frac{400}{w} = 6 \times 10.05 + \frac{400}{3.17} = 60.3 + 126.18 = 186.48$$ **Final answers:** - Width $w \approx 3.17$ cm - Height $h \approx 4.97$ cm - Minimum surface area $A \approx 186.48$ cm²