Let's break down the problem and solve it step by step. --- ## **(a) Problem Formulation** ### **1. Decision Variables** Let: - \( L \): **Outside length** of the tank (in ft) - \( W \): **Outside width** of the tank (in ft) ### **2. Objective Function** Minimize the **total cost** of concrete construction. #### **Calculation of Volume of Concrete:** - **Outside Length = L** - **Outside Width = W** - **Outside Depth = 10 ft** (inside) + 2 × 1 ft (walls) = **12 ft** (outside) - **Thickness = 1 ft** ##### **a. Volume of Concrete for Walls:** - **Side Walls (2 pieces):** - Height = 10 ft (inside), thickness = 1 ft, length = L (outside) - Each wall: \( L \times 10 \times 1 \) - Total for both: \( 2 \times L \times 10 \times 1 \) - **End Walls (2 pieces):** - Height = 10 ft, thickness = 1 ft, width = W (outside) - Each wall: \( W \times 10 \times 1 \) - Total for both: \( 2 \times W \times 10 \times 1 \) ##### **b. Volume of Concrete for Base and Cover:** - **Base:** \( L \times W \times 1 \) - **Cover:** \( L \times W \times 1 \) ##### **c. Total Volume:** \[ \text{Total Volume} = 2L \cdot 10 \cdot 1 + 2W \cdot 10 \cdot 1 + 2L \cdot W \cdot 1 \] \[ = 20L + 20W + 2LW \quad \text{(in ft}^3\text{)} \] Convert to \( m^3 \): \[ \text{Total Volume (in } m^3) = \frac{20L + 20W + 2LW}{35.315} \] ##### **d. Cost Function:** \[ \text{Cost} = \frac{20L + 20W + 2LW}{35.315} \times 550 \] --- ### **3. Constraints** #### **a. Area Constraint:** \[ L \times W \leq 600 \quad \text{(ft}^2\text{)} \] #### **b. Length and Width Relationship:** \[ L \geq 2W \] #### **c. Width Limitation:** \[ W \leq 20 \] #### **d. Positive Dimensions:** \[ L > 0,\ W > 0 \] --- ### **Summary (Formulation)** - **Decision Variables:** \( L, W \) - **Objective:** Minimize \(\quad C = \frac{20L + 20W + 2LW}{35.315} \times 550 \) - **Subject to:** - \( L \times W \leq 600 \) - \( L \geq 2W \) - \( W \leq 20 \) - \( L > 0, W > 0 \) --- ## **(b) Solve Using Excel Solver** ### **How to Set Up in Excel (Step-by-Step):** 1. **Set up cells:** - B1: `L` (Length in ft) - B2: `W` (Width in ft) - B3: `=20*B1+20*B2+2*B1*B2` (Total concrete volume in ft³) - B4: `=B3/35.315` (Total concrete volume in m³) - B5: `=B4*550` (Total cost in $) - B6: `=B1*B2` (Area) - B7: `=B1-2*B2` (L - 2W) 2. **Set initial values:** - B1: 40 (example) - B2: 15 (example) 3. **Open Solver:** - **Set Objective:** B5 (Minimize) - **By Changing Variable Cells:** B1, B2 - **Subject to Constraints:** - B6 <= 600 - B1 >= 2*B2 (or B7 >= 0) - B2 <= 20 - B1 > 0, B2 > 0 - **Solving Method:** GRG Nonlinear 4. **Solve.** 5. **Round answers to nearest 3":** (3 inches = 0.25 ft). --- ### **Sample Calculation (Manual Trial for Understanding):** Let’s try \( W = 20 \) (max allowed): - \( L = \frac{600}{20} = 30 \) ft (to use all 600 ft²) - Check \( L \geq 2W \): \( 30 \geq 40 \)? **No**. Not feasible. Try \( W = 15 \): - \( L = \frac{600}{15} = 40 \) ft - \( L \geq 2 \times 15 = 30 \), OK. Compute: - Volume: \( 20 \times 40 + 20 \times 15 + 2 \times 40 \times 15 = 800 + 300 + 1200 = 2300 \) ft³ - \( 2300/35.315 = 65.15 \) m³ - Cost: \( 65.15 \times 550 = \$35,832.5 \) Try \( W = 20 \) and \( L = 40 \): - Area: \( 20 \times 40 = 800 \) ft² (too much, so not feasible) Try \( W = 13 \): - \( L = 600 / 13 = 46.15 \) ft - \( L \geq 26 \), OK. Check cost: - Volume: \( 20 \times 46.15 + 20 \times 13 + 2 \times 46.15 \times 13 = 923 + 260 + 1199.9 = 2382.9 \) - \( 2382.9/35.315 = 67.48 \) m³ - Cost: \( 67.48 \times 550 = \$37,114 \) **Lower width gives lower cost**. --- ### **Optimal Solution (Using Solver):** Run Solver in Excel as described above. The minimum will likely occur at the smallest width where \( L = 2W \) and \( L \times W = 600 \): Let \( L = 2W \), so \( (2W) \cdot W = 600 \implies W^2 = 300 \implies W = \sqrt{300} = 17.32 \) ft, \( L = 34.64 \) ft Round to nearest 3": - 17.32 ft ≈ 17.25 ft (17 ft 3"), - 34.64 ft ≈ 34.5 ft (34 ft 6") --- #### **Final Answer (Rounded to Nearest 3")** \[ \boxed{ \begin{array}{l} \textbf{Optimal Dimensions:} \\ \quad \text{Width (W)} = 17.25 \text{ ft} \\ \quad \text{Length (L)} = 34.5 \text{ ft} \\ \\ \textbf{Minimum Cost:} \\ \quad \text{Concrete Volume: } 2 \times 17.25 \times 34.5 + 20 \times 17.25 + 20 \times 34.5 = \\ \quad 2 \times 595.125 + 345 + 690 = 1190.25 + 345 + 690 = 2225.25 \text{ ft}^3 \\ \quad 2225.25/35.315 = 63.05 \text{ m}^3 \\ \quad \text{Cost} = 63.05 \times 550 = \$34,677.5 \\ \\ \text{(Rounded to nearest \$0.50)} \end{array} } \] --- ### **Summary Table** | Variable | Value (ft) | Rounded (nearest 3") | |----------|------------|----------------------| | Width (W) | 17.32 | **17.25** | | Length (L) | 34.64 | **34.5** | | Volume (ft³) | Volume (m³) | Cost (\$) | |--------------|-------------|-----------| | 2225.25 | 63.05 | 34,677.5 | --- ### **Excel Setup Screenshot (for documentation)** If you need a screenshot or template, let me know! --- **Let me know if you need the Excel file or further clarifications!**