# Step-by-Step Solution for HW3 Problem 1 ## **Step 1: Define Variables** Let: - \( x_1 \) = number of Model 3 cars produced - \( x_2 \) = number of Model S cars produced - \( x_3 \) = number of Model X cars produced ## **Step 2: Write the Objective Function** Profits are: - Model 3: \$35,000 per unit - Model S: \$85,000 per unit - Model X: \$90,000 per unit **Objective:** Maximize profit \( Z \): \[ Z = 35,000x_1 + 85,000x_2 + 90,000x_3 \] ## **Step 3: List the Constraints** ### **a. Daily demand cannot exceed:** - Model 3: 225 - Model S: 125 - Model X: 112 So, \[ x_1 \leq 225 \] \[ x_2 \leq 125 \] \[ x_3 \leq 112 \] ### **b. Contractual obligation:** At least 200 Model 3, 125 Model S, 100 Model X must be produced: \[ x_1 \geq 200 \] \[ x_2 \geq 125 \] \[ x_3 \geq 100 \] ### **c. Marketing constraint (at most 1,400 units in total):** \[ x_1 + x_2 + x_3 \leq 140 \] ### **d. Combined constraint for contract and marketing:** As per the hint, the last two constraints (contractual obligation and marketing) are combined: \[ x_1 + x_2 + x_3 \leq 140 \] \[ x_1 \geq 200 \] \[ x_2 \geq 125 \] \[ x_3 \geq 100 \] ### **e. Non-negativity:** \[ x_1, x_2, x_3 \geq \] --- ## **Step 4: Formulate the Linear Program** **Maximize:** \[ Z = 35,000x_1 + 85,000x_2 + 90,000x_3 \] **Subject to:** \[ 200 \leq x_1 \leq 225 \] \[ 125 \leq x_2 \leq 125 \] \[ 100 \leq x_3 \leq 112 \] \[ x_1 + x_2 + x_3 \leq 140 \] \[ x_1, x_2, x_3 \geq \] --- ## **Step 5: Find the Feasible Solution** From constraints: - \(x_1\) must be between 200 and 225 - \(x_2\) must be exactly 125 - \(x_3\) must be between 100 and 112 To maximize profit, maximize the most profitable car (Model X at \$90,000), subject to the constraints. - Set \(x_2 = 125\) (as it's both the min and max) - Set \(x_3 = 112\) (max possible) - Set \(x_1 = 225\) (max possible) Check if \(x_1 + x_2 + x_3 \leq 140\): \[ 225 + 125 + 112 = 462 \leq 140 \] Feasible. --- ### **Profit Calculation** \[ Z = 35,000 \times 225 + 85,000 \times 125 + 90,000 \times 112 \] \[ = 7,875,000 + 10,625,000 + 10,080,000 = \$28,580,000 \] --- ## **Step 6: Sensitivity Report & Shadow Prices** - **Reduced Cost:** The amount by which the objective function coefficient of a variable must improve before it becomes optimal to increase that variable from zero. - **Shadow Price:** The rate at which the objective function would increase if the right-hand side of a constraint is increased by one unit (while all other data remains constant). For this problem: - All variables are at their upper/lower bounds, so reduced costs are zero. - The shadow price for the marketing constraint is zero (since it is not binding; sum is much less than 140). --- ## **Step 7: Changes in Demand for Model 3** - **If minimum demand for Model 3 increases to 475:** Not possible, since the upper bound is 225. So, the problem becomes infeasible. - **If minimum demand for Model 3 increases to 476:** Also infeasible for the same reason. --- ## **Final Summary** ### **Optimal Production Plan:** - Produce **225 units of Model 3** - Produce **125 units of Model S** - Produce **112 units of Model X** ### **Maximum Profit:** - **\$28,580,000** ### **Sensitivity Analysis:** - **Reduced Costs:** All zero, as all variables are at their bounds. - **Shadow Prices:** Zero for marketing constraint (not binding). ### **Changes in Model 3 Demand:** - If the minimum required units for Model 3 increase to 475 or 476, the problem has **no feasible solution** due to the upper limit of 225 units. --- **If you need more details for each part (b), (c), (d), and (e), let me know!**