Let's break down the problem step by step: ### **Step 1: Define the Variables** Let: - Width of the base = \( x \) (in meters) - Length of the base = \( 2x \) (since it's twice the width) - Height = \( h \) (in meters) ### **Step 2: Volume Constraint** The volume is given as \( 8 \) m³: \[ \text{Volume} = \text{length} \times \text{width} \times \text{height} \] \[ 8 = 2x \cdot x \cdot h = 2x^2 h \] \[ h = \frac{8}{2x^2} = \frac{4}{x^2} \] --- ### **Step 3: Surface Area for Cost Calculation** - **Base area** = \( 2x \times x = 2x^2 \) - **Side areas:** - Two sides of \( x \times h \): \( 2(xh) \) - Two sides of \( 2x \times h \): \( 2(2x h) = 4x h \) - **Total side area:** \( 2xh + 4xh = 6xh \) - **Open top:** No cost for the top --- ### **Step 4: Cost Function** - **Base cost:** \( 5 \) dollars per m² - **Side cost:** \( 10 \) dollars per m² \[ \text{Total Cost} = \text{Cost of base} + \text{Cost of sides} \] \[ = 5 \cdot (2x^2) + 10 \cdot (6xh) \] \[ = 10x^2 + 60xh \] Substitute \( h = \frac{4}{x^2} \): \[ \text{Total Cost} = 10x^2 + 60x \left(\frac{4}{x^2}\right) \] \[ = 10x^2 + \frac{240}{x} \] --- ### **Step 5: Minimize the Cost** Let \( C(x) = 10x^2 + \frac{240}{x} \) Take the derivative and set to zero: \[ C'(x) = 20x - \frac{240}{x^2} \] Set \( C'(x) = 0 \): \[ 20x - \frac{240}{x^2} = 0 \] \[ 20x = \frac{240}{x^2} \] \[ 20x^3 = 240 \] \[ x^3 = 12 \] \[ x = \sqrt[3]{12} \] \[ x \approx 2.29 \text{ meters} \] --- ### **Step 6: Verification (Second Derivative Test)** \[ C''(x) = 20 + \frac{480}{x^3} \] For \( x > 0 \), \( C''(x) > 0 \), so it's a minimum. --- ### **Step 7: Final Answers** **Width:** \( \boxed{x \approx 2.29 \text{ meters}} \) **Length:** \( 2x \approx 4.58 \text{ meters} \) **Height:** \( h = \frac{4}{x^2} \approx \frac{4}{(2.29)^2} \approx 0.764 \text{ meters} \) --- ### **Step 8: Minimum Cost** Plug \( x \) back into the cost function: \[ C(x) = 10x^2 + \frac{240}{x} \] \[ C(2.29) \approx 10(2.29)^2 + \frac{240}{2.29} \] \[ \approx 10(5.24) + 104.8 \approx 52.4 + 104.8 = 157.2 \text{ dollars} \] --- ## **Summary Table** | Dimension | Value (rounded) | |-------------------|------------------| | Width (\(x\)) | 2.29 m | | Length (\(2x\)) | 4.58 m | | Height (\(h\)) | 0.76 m | | Minimum Cost | \$157.20 | --- ### **Explanation** - The width that gives the cheapest material cost is \( \mathbf{x = \sqrt[3]{12} \approx 2.29} \) meters. - The cost is minimized because the second derivative is positive, confirming a minimum. --- **Let me know if you need a diagram or more explanation!**

Certainly! Let's go through the problem step-by-step with detailed calculations, explanations, and the final answers. --- ### **Problem Summary:** - You are constructing a box with an open top. - The volume of the box is **8 m³**. - The length \( L \) is **twice** the width \( W \): \( L = 2W \). - The cost: - Base: \$5 per m² - Sides: \$10 per m² - **Goal:** Minimize the total cost of materials. --- ### **Step 1: Define Variables** Let: - Width = \( W = x \) meters - Length = \( L = 2x \) meters - Height = \( H = h \) meters --- ### **Step 2: Write the Volume Constraint** \[ V = \text{length} \times \text{width} \times \text{height} \] \[ 8 = (2x) \times x \times h \] \[ 8 = 2x^2 h \] Solve for \( h \): \[ h = \frac{8}{2x^2} = \frac{4}{x^2} \] --- ### **Step 3: Surface Area for Cost Calculation** - **Base area:** \( A_{base} = L \times W = 2x \times x = 2x^2 \) - **Side areas:** - Two sides of \( W \times H \): \( 2 \times (x \times h) = 2xh \) - Two sides of \( L \times H \): \( 2 \times (2x \times h) = 4xh \) - **Total side area:** \[ A_{sides} = 2xh + 4xh = 6xh \] --- ### **Step 4: Write the Cost Function** - Cost of the base: \$5 per m² - Cost of sides: \$10 per m² \[ \text{Total Cost} = 5 \times A_{base} + 10 \times A_{sides} \] \[ = 5 \times 2x^2 + 10 \times 6xh \] \[ = 10x^2 + 60xh \] Substitute \( h = \frac{4}{x^2} \): \[ \text{Total Cost} = 10x^2 + 60x \times \frac{4}{x^2} \] \[ = 10x^2 + \frac{240x}{x^2} \] \[ = 10x^2 + \frac{240}{x} \] Define the cost function: \[ C(x) = 10x^2 + \frac{240}{x} \] --- ### **Step 5: Minimize the Cost Function** Differentiate \( C(x) \) with respect to \( x \): \[ C'(x) = 20x - \frac{240}{x^2} \] Set derivative to zero for minimization: \[ 20x - \frac{240}{x^2} = 0 \] Multiply through by \( x^2 \) to clear denominator: \[ 20x^3 - 240 = 0 \] Solve for \( x^3 \): \[ 20x^3 = 240 \] \[ x^3 = \frac{240}{20} = 12 \] \[ x = \sqrt[3]{12} \] Calculate \( \sqrt[3]{12} \): \[ x \approx 2.289 \text{ meters} \] --- ### **Step 6: Verify that this is a minimum** Second derivative: \[ C''(x) = 20 + \frac{480}{x^3} \] Since \( x > 0 \), \( C''(x) > 0 \), confirming a minimum at \( x \approx 2.289 \). --- ### **Step 7: Calculate the corresponding dimensions** - Width: \[ W = x \approx 2.289 \text{ meters} \] - Length: \[ L = 2x \approx 2 \times 2.289 = 4.578 \text{ meters} \] - Height: \[ h = \frac{4}{x^2} = \frac{4}{(2.289)^2} \approx \frac{4}{5.236} \approx 0.763 \text{ meters} \] --- ### **Step 8: Calculate the minimum cost** Plug \( x \approx 2.289 \) into the cost function: \[ C(x) = 10x^2 + \frac{240}{x} \] Compute: \[ 10 \times (2.289)^2 \approx 10 \times 5.236 \approx 52.36 \] \[ \frac{240}{2.289} \approx 104.8 \] Total: \[ C \approx 52.36 + 104.8 \approx \boxed{\$157.16} \] --- ### **### **Final Answers:** | Dimension | Approximate Value | |--------------------------------|---------------------------------------| | Width \( W \) | **2.29 meters** | | Length \( L = 2W \) | **4.58 meters** | | Height \( H \) | **0.76 meters** | | Minimum Cost | **\$157.16** | --- **Summary:** To minimize the material cost, construct the box with: - Width ≈ 2.29 m - Length ≈ 4.58 m - Height ≈ 0.76 m The total minimum cost is approximately **\$157.16**. --- Let me know if you'd like any further clarifications!