1. **State the problem:** We want to find the dimensions of a box with base length $l$ and base width $w$ such that $l=6w$, volume $V=20$ in$^3$, and the cost of materials is minimized. 2. **Given:** - $l=6w$ - Volume $V = lwh = 20$ - Cost per unit area for sides = 3 - Cost per unit area for top and bottom = 15 3. **Express height $h$ in terms of $w$: ** $$V = lwh = 6w \cdot w \cdot h = 6w^2 h = 20 \implies h = \frac{20}{6w^2} = \frac{10}{3w^2}$$ 4. **Surface areas:** - Sides area = 2(height × length) + 2(height × width) = $2lh + 2wh$ - Top and bottom area = 2 × (length × width) = $2lw$ 5. **Cost function $C(w)$:** $$C = 3 \times \text{sides area} + 15 \times \text{top and bottom area} = 3(2lh + 2wh) + 15(2lw)$$ Substitute $l=6w$: $$C = 3(2 \cdot 6w \cdot h + 2w \cdot h) + 15(2 \cdot 6w \cdot w) = 3(12wh + 2wh) + 15(12w^2) = 3(14wh) + 180w^2 = 42wh + 180w^2$$ 6. Substitute $h = \frac{10}{3w^2}$: $$C = 42w \cdot \frac{10}{3w^2} + 180w^2 = 42 \cdot \frac{10}{3w} + 180w^2 = \frac{420}{3w} + 180w^2 = \frac{140}{w} + 180w^2$$ 7. **Minimize $C(w)$:** Take derivative and set to zero: $$C'(w) = -\frac{140}{w^2} + 360w = 0$$ 8. Solve for $w$: $$360w = \frac{140}{w^2} \implies 360w^3 = 140 \implies w^3 = \frac{140}{360} = \frac{7}{18}$$ $$w = \sqrt[3]{\frac{7}{18}} \approx 0.5299$$ 9. Calculate $l$ and $h$: $$l = 6w = 6 \times 0.5299 = 3.1794$$ $$h = \frac{10}{3w^2} = \frac{10}{3 \times (0.5299)^2} \approx \frac{10}{3 \times 0.2808} = \frac{10}{0.8424} \approx 11.87$$ 10. Check options: The closest to $w=0.5299$, $l=3.1794$, $h=11.87$ is option b but $h$ differs. Recalculate $l$ carefully: $$l = 6 \times 0.5299 = 3.1794$$ Option b has $l=2.3794$ and $h=4.2568$, so no match. Re-examine step 9: The volume is $20 = lwh = 6w \cdot w \cdot h = 6w^2 h$, so $$h = \frac{20}{6w^2} = \frac{10}{3w^2}$$ Calculate $h$ again with $w=0.5299$: $$h = \frac{10}{3 \times (0.5299)^2} = \frac{10}{3 \times 0.2808} = \frac{10}{0.8424} = 11.87$$ This is much larger than any $h$ in options. Check if cost function or problem interpretation is correct. Since options have smaller $h$, check if $l=6w$ or $l=6 \times w$ is correct. Given $l=6w$, volume $20 = lwh = 6w \cdot w \cdot h = 6w^2 h$, so $h = \frac{20}{6w^2}$. Try option b values: $$w=0.5299, l=2.3794, h=4.2568$$ Check if $l=6w$: $$6 \times 0.5299 = 3.1794 \neq 2.3794$$ No. Try option a: $$w=0.6299, l=3.3794, h=5.2568$$ $$6 \times 0.6299 = 3.7794 \neq 3.3794$$ No. Try option d: $$w=0.7299, l=4.3794, h=6.2568$$ $$6 \times 0.7299 = 4.3794$$ Yes, matches. Check volume: $$V = lwh = 4.3794 \times 0.7299 \times 6.2568 \approx 20$$ Calculate cost for option d: Sides area: $$2lh + 2wh = 2 \times 4.3794 \times 6.2568 + 2 \times 0.7299 \times 6.2568 = 54.82 + 9.13 = 63.95$$ Top and bottom area: $$2lw = 2 \times 4.3794 \times 0.7299 = 6.39$$ Cost: $$3 \times 63.95 + 15 \times 6.39 = 191.85 + 95.85 = 287.7$$ Check option b cost: $$l=2.3794, w=0.5299, h=4.2568$$ Sides area: $$2lh + 2wh = 2 \times 2.3794 \times 4.2568 + 2 \times 0.5299 \times 4.2568 = 20.25 + 4.51 = 24.76$$ Top and bottom area: $$2lw = 2 \times 2.3794 \times 0.5299 = 2.52$$ Cost: $$3 \times 24.76 + 15 \times 2.52 = 74.28 + 37.8 = 112.08$$ Option b has lower cost, but $l \neq 6w$. Therefore, the only option satisfying $l=6w$ is d. **Final answer:** Option d: $w=0.7299$, $l=4.3794$, $h=6.2568$.