1. **Problem Statement:** A closed cuboid made of paper has a volume of 150 cm³. The height is $h$, the width is $w$, and the length is triple the width, i.e., $3w$. 2. **Diagram Description:** The cuboid has dimensions: height = $h$ cm, width = $w$ cm, length = $3w$ cm. 3. **Express height $h$ in terms of $w$:** Volume formula for cuboid: $$V = \text{length} \times \text{width} \times \text{height}$$ Given volume: $$150 = 3w \times w \times h = 3w^2 h$$ Solve for $h$: $$h = \frac{150}{3w^2} = \frac{150}{3w^2}$$ Intermediate step showing cancellation: $$h = \frac{150}{\cancel{3}w^2} = \frac{50}{w^2}$$ 4. **Express surface area $A$ in terms of $w$:** Surface area of closed cuboid: $$A = 2(\text{length} \times \text{width} + \text{width} \times \text{height} + \text{length} \times \text{height})$$ Substitute length = $3w$, width = $w$, height = $h$: $$A = 2(3w \times w + w \times h + 3w \times h) = 2(3w^2 + w h + 3w h) = 2(3w^2 + 4w h)$$ Substitute $h = \frac{50}{w^2}$: $$A = 2\left(3w^2 + 4w \times \frac{50}{w^2}\right) = 2\left(3w^2 + \frac{200}{w}\right) = 6w^2 + \frac{400}{w}$$ 5. **Minimize surface area $A$:** To find minimum, differentiate $A$ with respect to $w$ and set to zero: $$\frac{dA}{dw} = 12w - \frac{400}{w^2} = 0$$ Multiply both sides by $w^2$: $$w^2 \times 12w - w^2 \times \frac{400}{w^2} = 0 \Rightarrow 12w^3 - 400 = 0$$ Solve for $w^3$: $$12w^3 = 400 \Rightarrow w^3 = \frac{400}{12} = \frac{100}{3}$$ Find $w$: $$w = \sqrt[3]{\frac{100}{3}} \approx 3.17$$ 6. **Find corresponding $h$:** $$h = \frac{50}{w^2} = \frac{50}{(3.17)^2} = \frac{50}{10.05} \approx 4.97$$ 7. **Find minimum surface area $A$:** $$A = 6w^2 + \frac{400}{w} = 6 \times 10.05 + \frac{400}{3.17} = 60.3 + 126.18 = 186.48$$ **Final answers:** - $w \approx 3.17$ cm - $h \approx 4.97$ cm - Minimum surface area $A \approx 186.48$ cm²