1. **State the problem:** We have a rectangular solid spacer with dimensions length $L=8$ ft, height $H=14$ ft, and width $W=32$ ft. There are two cylindrical holes passing through the width, each with radius $r=2$ ft and length equal to the width $32$ ft. We need to find: - The cost of the plastic required to make one spacer. - The cost to paint all exposed surfaces including inside the holes. 2. **Formulas and rules:** - Volume of rectangular solid: $$V_{box} = L \times H \times W$$ - Volume of one cylinder: $$V_{cyl} = \pi r^2 h$$ where $h$ is the length of the cylinder. - Total volume of plastic: $$V_{plastic} = V_{box} - 2 \times V_{cyl}$$ - Surface area of rectangular solid: $$SA_{box} = 2(LH + HW + LW)$$ - Surface area of one cylindrical hole (lateral surface only, since holes pass through): $$SA_{cyl} = 2 \pi r h$$ - Total paint area: $$SA_{paint} = SA_{box} + 2 \times SA_{cyl}$$ 3. **Calculate volumes:** $$V_{box} = 8 \times 14 \times 32 = 3584 \text{ ft}^3$$ $$V_{cyl} = \pi \times 2^2 \times 32 = \pi \times 4 \times 32 = 128\pi \approx 402.12 \text{ ft}^3$$ $$V_{plastic} = 3584 - 2 \times 402.12 = 3584 - 804.24 = 2779.76 \text{ ft}^3$$ 4. **Calculate cost of plastic:** Cost per cubic foot = 3.15 $$\text{Cost}_{plastic} = 2779.76 \times 3.15 = 8753.30$$ 5. **Calculate surface areas:** $$SA_{box} = 2(8 \times 14 + 14 \times 32 + 8 \times 32) = 2(112 + 448 + 256) = 2(816) = 1632 \text{ ft}^2$$ $$SA_{cyl} = 2 \pi \times 2 \times 32 = 128\pi \approx 402.12 \text{ ft}^2$$ $$SA_{paint} = 1632 + 2 \times 402.12 = 1632 + 804.24 = 2436.24 \text{ ft}^2$$ 6. **Calculate cost of paint:** Cost per square foot = 0.15 $$\text{Cost}_{paint} = 2436.24 \times 0.15 = 365.44$$ **Final answers:** - Cost of plastic: 8753.30 - Cost to paint: 365.44