Let's answer each subquestion step by step using the data from the provided table. --- ## a) Distance between Ogden and Lubbock **Formula:** Distance = \(\sqrt{(X_2 - X_1)^2 + (Y_2 - Y_1)^2}\) **Coordinates:** - Ogden: (250, 600) - Lubbock: (450, 300) **Calculation:** \[ \text{Distance} = \sqrt{(450 - 250)^2 + (300 - 600)^2} \] \[ = \sqrt{(200)^2 + (-300)^2} \] \[ = \sqrt{40000 + 90000} \] \[ = \sqrt{130000} \] \[ = 360.55 \text{ miles (rounded to two decimal places)} \] --- ## b) Cost to transport 400 tons from Springfield to Seattle **Springfield coordinates:** (225, 825) **Seattle coordinates:** (50, 1200) **Transportation cost:** \$1.4/ton-mile (Springfield's rate) **Step 1: Find the distance** \[ \text{Distance} = \sqrt{(50 - 225)^2 + (1200 - 825)^2} \] \[ = \sqrt{(-175)^2 + (375)^2} \] \[ = \sqrt{30625 + 140625} \] \[ = \sqrt{171250} \] \[ = 413.75 \text{ miles} \] **Step 2: Calculate transport cost** \[ \text{Total cost} = \text{tons} \times \text{distance} \times \text{cost per ton-mile} \] \[ = 400 \times 413.75 \times 1.4 \] \[ = 400 \times 579.25 \] \[ = \$231,700 \] --- ## c) Optimal location for the Super Assembly Center The optimal location is the **center of gravity** (weighted average based on supply): \[ X_{opt} = \frac{\sum D_n X_n}{\sum D_n} \] \[ Y_{opt} = \frac{\sum D_n Y_n}{\sum D_n} \] **Supply data:** - North Haverbrook: 650 tons, (700, 1200) - Ogden: 450 tons, (250, 600) - Springfield: 400 tons, (225, 825) **Total supply:** 650 + 450 + 400 = 1500 tons **Calculate \(X_{opt}\):** \[ X_{opt} = \frac{650 \times 700 + 450 \times 250 + 400 \times 225}{1500} \] \[ = \frac{455000 + 112500 + 90000}{1500} \] \[ = \frac{657500}{1500} = 438.33 \] **Calculate \(Y_{opt}\):** \[ Y_{opt} = \frac{650 \times 1200 + 450 \times 600 + 400 \times 825}{1500} \] \[ = \frac{780000 + 270000 + 330000}{1500} \] \[ = \frac{1380000}{1500} = 920 \] **Optimal Location:** \[ \boxed{(438.33,\ 920)} \] --- ## d) Minimized shipping cost that satisfies all market demand Assume the super assembly center is at the optimal location \((438.33, 920)\). **Market demand and locations:** - Seattle: 300 tons, (50, 1200) - Lubbock: 200 tons, (450, 300) - Nashville: 500 tons, (800, 250) - Philadelphia: 500 tons, (950, 1100) **Transportation cost \((F_n)\):** \$2.25/ton-mile for all markets ### Step 1: Calculate distances #### Seattle: \[ \text{Distance} = \sqrt{(438.33-50)^2 + (920-1200)^2} \] \[ = \sqrt{151,094 + 78,400} = \sqrt{229,494} = 479.14 \] #### Lubbock: \[ \text{Distance} = \sqrt{(438.33-450)^2 + (920-300)^2} \] \[ = \sqrt{(11.67)^2 + (620)^2} = \sqrt{136.23 + 384,400} = \sqrt{384,536.23} = 620.04 \] #### Nashville: \[ \text{Distance} = \sqrt{(438.33-800)^2 + (920-250)^2} \] \[ = \sqrt{(361.67)^2 + (670)^2} = \sqrt{130,806.39 + 448,900} = \sqrt{579,706.39} = 761.45 \] #### Philadelphia: \[ \text{Distance} = \sqrt{(438.33-950)^2 + (920-1100)^2} \] \[ = \sqrt{(511.67)^2 + (180)^2} = \sqrt{261,828.19 + 32,400} = \sqrt{294,228.19} = 542.44 \] ### Step 2: Calculate shipping cost for each \[ \text{Cost} = \text{tons} \times \text{distance} \times 2.25 \] - **Seattle:** \(300 \times 479.14 \times 2.25 = 323,440.5\) - **Lubbock:** \(200 \times 620.04 \times 2.25 = 279,018\) - **Nashville:** \(500 \times 761.45 \times 2.25 = 856,631.25\) - **Philadelphia:** \(500 \times 542.44 \times 2.25 = 609,245\) **Total Minimized Shipping Cost:** \[ = 323,440.5 + 279,018 + 856,631.25 + 609,245 = \boxed{2,068,335.75} \] --- ## **Final Summary** **a) Distance between Ogden and Lubbock:** \[ \boxed{360.55 \text{ miles}} \] **b) Cost to transport 400 tons from Springfield to Seattle:** \[ \boxed{\$231,700} \] **c) Optimal location for the Super Assembly Center:** \[ \boxed{(438.33,\ 920)} \] **d) Minimized shipping cost (total):** \[ \boxed{\$2,068,335.75} \] All calculations are shown step-by-step above. If you need Excel formulas or output, let me know!