\]Liam is designing a cylindrical water tank that must hold exactly 500 liters (equivalent to 0.5 m³) of water. If the material cost for the curved lateral surface is $40 per m² and the top and bottom each cost $55 per m² due to thicker reinforcement, what is the minimum cost to construct the tank, minimizing the surface area?

["Title: How to Minimize the Cost of a 500-Liter Cylindrical Water Tank: Optimizing Surface Area and Material Costs", "---", "Designing a cylindrical water tank requires more than just choosing the right diameter and height—it’s also critical to minimize construction costs while ensuring the tank holds exactly 500 liters (0.5 m³) of water. In this article, we explore how Liam can optimize his cylindrical tank design by minimizing total material costs through surface area efficiency.", "### Understanding the Geometry and Requirements", "The tank must hold 0.5 m³ of water. For a cylinder, volume is given by:", "[\nV = \pi r^2 h\n]", "Where:\n- ( r ) = radius (in meters)\n- ( h ) = height (in meters)", "Given:\n[\nV = 0.5~\ ext{m}^3\n]\nSo:\n[\n\pi r^2 h = 0.5 \quad \Rightarrow \quad h = \frac{0.5}{\pi r^2}\n]", "The total surface area of the cylindrical tank includes:\n- Curved lateral surface area: ( A_{\ ext{lat}} = 2\pi r h ), costing $40 per m²\n- Top and bottom areas: each is a circle with area ( \pi r^2 ), costing $55 per m²", "Total cost function:\n[\nC = 40(2\pi r h) + 55(2\pi r^2) = 80\pi r h + 110\pi r^2\n]", "Substitute ( h = \frac{0.5}{\pi r^2} ) into the cost equation:", "[\nC(r) = 80\pi r \left( \frac{0.5}{\pi r^2} \right) + 110\pi r^2 = \frac{40}{r} + 110\pi r^2\n]", "Our goal is to minimize ( C(r) = \frac{40}{r} + 110\pi r^2 ) for ( r > 0 ).", "---", "### Minimizing the Cost Function", "To find the minimum, take the derivative of ( C(r) ) with respect to ( r ):", "[\nC'(r) = -\frac{40}{r^2} + 220\pi r\n]", "Set ( C'(r) = 0 ) for critical points:", "[\n-\frac{40}{r^2} + 220\pi r = 0 \quad \Rightarrow \quad 220\pi r = \frac{40}{r^2}\n]", "[\n220\pi r^3 = 40 \quad \Rightarrow \quad r^3 = \frac{40}{220\pi} = \frac{2}{11\pi}\n]", "[\nr = \left( \frac{2}{11\pi} \right)^{1/3}\n]", "Now calculate numerical value:", "[\nr \approx \left( \frac{2}{34.557} \right)^{1/3} \approx (0.0579)^{1/3} \approx 0.388~\ ext{m}\n]", "Use this optimized radius to find corresponding height:", "[\nh = \frac{0.5}{\pi r^2} = \frac{0.5}{\pi (0.388)^2} \approx \frac{0.5}{\pi \cdot 0.150} \approx \frac{0.5}{0.471} \approx 1.062~\ ext{m}\n]", "---", "### Calculate Minimum Surface Area and Cost", "Use exact expressions to maintain accuracy. Recall:", "[\nC(r) = \frac{40}{r} + 110\pi r^2\n]", "Substitute ( r^3 = \frac{2}{11\pi} \Rightarrow r = \left( \frac{2}{11\pi} \right)^{1/3} )", "Then:", "[\n\frac{1}{r} = \left( \frac{11\pi}{2} \right)^{1/3}, \quad r^2 = \left( \frac{2}{11\pi} \right)^{2/3}\n]", "So:", "[\nC = 40 \left( \frac{11\pi}{2} \right)^{1/3} + 110\pi \left( \frac{2}{11\pi} \right)^{2/3}\n]", "Factor out ( \left( \frac{2}{11\pi} \right)^{1/3} ):", "[\nC = \left( \frac{2}{11\pi} \right)^{1/3} \left[ 40 \left( \frac{11\pi}{2} \right)^{1/3} \cdot \left( \frac{11\pi}{2} \right)^{-1/3} \cdot (11\pi) + 110\pi \cdot \left( \frac{2}{11\pi} \right) \right]\n]", "Actually, a simpler way: since we know ( r^3 = \frac{2}{11\pi} ), then:", "[\n\frac{40}{r} = 40 r^{-1} = 40 \left( \frac{11\pi}{2} \right)^{1/3}, \quad 110\pi r^2 = 110\pi \left( \frac{2}{11\pi} \right)^{2/3}\n]", "But numerically is more practical for real-world estimation:", "Using ( r \approx 0.388~\ ext{m} ):", "[\n\frac{40}{r} \approx \frac{40}{0.388} \approx 103.09\n]\n[\n110\pi r^2 \approx 110 \cdot 3.1416 \cdot (0.388)^2 \approx 345.576 \cdot 0.1505 \approx 52.04\n]\n[\nC_{\ ext{min}} \approx 103.09 + 52.04 = 155.13~\ ext{dollars}\n]", "Thus, the minimum construction cost is approximately $155.13, achieved with optimal radius and height calculated above.", "---", "### Conclusion", "By minimizing surface area through calculus-based optimization, Liam ensures the cylindrical tank holds 500 liters at the lowest possible cost. The key insight is balancing cost-effective lateral reinforcement with specialized top/bottom materials, showing that smart mathematical modeling directly reduces real-world expenses.", "Investing time to calculate optimal dimensions leads to significant savings—especially in large-scale projects where material use impacts overall budget. For precise engineering, always verify critical dimensions before fabrication.", "---", "Keywords: cylindrical water tank cost optimization, minimize cylinder surface area, water storage tank design, mathematical optimization in engineering, cylinder geometry cost calculation, 500-liter tank surface area", "Meta Description: Learn how Liam minimizes construction cost for a 500L cylindrical water tank by optimizing radius and height using calculus, achieving the lowest surface area and material expense."]