A rectangular box with no top is to be made with a square base and a volume of 108 cubic meters. If the material for the base costs twice as much per square meter as the sides, what is the minimum possible cost surface area? (Assume cost per unit area for sides is 1 unit)

["Optimizing the Cost of a Rectangular Box with a Square Base and No Top: Minimizing Surface Area Cost", "In engineering, architecture, and manufacturing, designing lightweight yet cost-efficient boxes is crucial. This article explores a classic optimization problem: constructing a rectangular box with a square base and no top, having a volume of 108 cubic meters, where the base material costs twice as much per square meter as the side panels. The goal is to minimize the total surface cost under these constraints.", "---", "### Problem Setup", "We are given:\n- A box with a square base → base and top have identical square dimensions.\n- The box has no top, so the surface consists only of the base and the four vertical side panels.\n- Volume: ( V = 108 ) cubic meters.\n- Let the side length of the square base be ( x ) meters.\n- Let the height (depth) of the box be ( h ) meters.\n- Cost per square meter:\n - Base and sides: $1 per m².\n - Base material costs twice as much: $2 per m².", "We aim to minimize the total cost surface area, i.e., minimize total cost based on the varying material prices.", "---", "### Step 1: Express Volume and Surface Area", "Volume Equation:\n[\nV = x^2 h = 108 \implies h = \frac{108}{x^2}\n]", "Surface Area Components:\n- Base area: ( x^2 ) (costs $2 per m²)\n- Four sides: each of area ( x \ imes h ), total side area = ( 4xh ) (each $1 per m²)", "Total Cost ( C ):\n[\nC = (\ ext{Base cost}) + (\ ext{Sides cost}) = 2x^2 + 1 \cdot (4xh) = 2x^2 + 4xh\n]", "Substitute ( h = \frac{108}{x^2} ):\n[\nC = 2x^2 + 4x \left( \frac{108}{x^2} \right) = 2x^2 + \frac{432}{x}\n]", "---", "### Step 2: Minimize Cost Function", "We now minimize ( C(x) = 2x^2 + \frac{432}{x} ) for ( x > 0 ).", "Take derivative:\n[\n\frac{dC}{dx} = 4x - \frac{432}{x^2}\n]", "Set derivative to zero for critical points:\n[\n4x - \frac{432}{x^2} = 0 \implies 4x = \frac{432}{x^2} \implies 4x^3 = 432 \implies x^3 = 108 \implies x = \sqrt[3]{108}\n]", "Simplify ( \sqrt[3]{108} ):\n[\n108 = 27 \ imes 4 \implies x = \sqrt[3]{27 \cdot 4} = 3\sqrt[3]{4}\n]", "Now compute corresponding height:\n[\nh = \frac{108}{x^2} = \frac{108}{(3\sqrt[3]{4})^2} = \frac{108}{9 \cdot \sqrt[3]{16}} = \frac{12}{\sqrt[3]{16}} = \frac{12}{2\sqrt[3]{2}} = \frac{6}{\sqrt[3]{2}}\n]", "But we don’t need ( h ) explicitly; the minimized cost is:\n[\nC_{\ ext{min}} = 2x^2 + \frac{432}{x} = 2(9 \cdot \sqrt[3]{16}) + \frac{432}{3\sqrt[3]{4}} = 18 \cdot \sqrt[3]{16} + 144 \cdot \sqrt[3]{[1/4]} \n]", "Alternatively, plug ( x^3 = 108 ) into cost expression:", "Let ( x = \sqrt[3]{108} ), then:\n[\nC = 2(\sqrt[3]{108})^2 + \frac{432}{\sqrt[3]{108}} = 2 \cdot 108^{2/3} + 432 \cdot 108^{-1/3}\n]", "Note: ( 108^{2/3} = (108^{1/3})^2 ), let ( y = 108^{1/3} ), so:\n[\nC = 2y^2 + \frac{432}{y} = 2y^2 + 432 y^{-1}\n]\nBut from earlier: ( y^3 = 108 ), so ( y^2 = 108 / y )", "Substitute:\n[\nC = 2(108 / y) + 432 / y = \frac{216}{y} + \frac{432}{y} = \frac{648}{y} = \frac{648}{108^{1/3}}\n]", "Now ( 648 = 6 \cdot 108 ), so:\n[\nC = \frac{6 \cdot 108}{108^{1/3}} = 6 \cdot 108^{2/3}\n]", "But simpler: ( 108^{1/3} = (27 \cdot 4)^{1/3} = 3 \cdot 4^{1/3} ), so:\n[\n108^{2/3} = (3^3 \cdot 4^{1})^^{2/3} = 3^2 \cdot 4^{2/3} = 9 \cdot 4^{2/3}\n]\nThus:\n[\nC = 6 \cdot 9 \cdot 4^{2/3} = 54 \cdot 4^{2/3}\n]", "Alternatively, accept simplest form:", "[\nC_{\ ext{min}} = 6 \cdot 108^{2/3}\n]", "But compute numerically for clarity:\n[\n108^{1/3} \approx 4.762\n]\n[\n108^{2/3} \approx (4.762)^2 \approx 22.68\n]\n[\nC \approx 6 \cdot 22.68 \approx 136.08\n]", "But exact minimal cost is elegant to express algebraically.", "---", "### Step 3: Confirm Minimum Using Second Derivative", "[\n\frac{d^2C}{dx^2} = 4 + \frac{864}{x^3}\n]\nFor ( x > 0 ), this is always positive → minimum at critical point.", "---", "### Key Insight: Cost Function Depends on Volume", "Interestingly, the minimization yields a unique minimal cost regardless of expensive material heterogeneity. However, the trade-off between base (costlier) and sides (cheaper) is resolved through optimal proportions.", "In this optimal box:\n- ( x = \sqrt[3]{108} \approx 4.762 , \ ext{m} )\n- ( h = \frac{108}{x^2} = \frac{108}{(\sqrt[3]{108})^2} = \sqrt[3]{108} \approx 4.762 , \ ext{m} )", "Note: Height equals base side — a surprising symmetry confirming optimal material use.", "---", "### Final Answer: Minimum Cost Surface Area", "The minimum possible cost surface area for the box, incorporating the higher cost of the base, is:", "[\n\boxed{6 \cdot 108^{2/3}} \quad \ ext{(in cost units)}\n]", "This value is approximately $136.07, but the exact form represents the precise cost for minimum expense under the given constraints.", "---", "### Practical Tip\nWhen designing such containers, balance competing material costs. Pre-calculating via calculus ensures efficient use of resources — saving money without sacrificing structural volume.", "Optimize with both geometry and economics in mind. A square-base, no-top box of 108 m³ with base cost doubled achieves minimum cost surface area efficiently — a prime example of applied optimization in real-world design."]