Let's solve part (a) and (b) **step by step** with explanations at each step. --- ## **Step 1: Draw the Feasible Set (part a)** ### **Constraints:** 1. \(2x_1 + 3x_2 \leq 6\) 2. \(x_1 + 4x_2 \leq 5\) 3. \(x_1 \geq 0\) 4. \(x_2 \geq 0\) ### **Explanation:** - The feasible set is the region in the \(x_1, x_2\) plane that satisfies **all** the inequalities above. - Plot the lines: - \(2x_1 + 3x_2 = 6\) - \(x_1 + 4x_2 = 5\) - Shade **below** each line (since \(\leq\)). - Restrict to the **first quadrant** (\(x_1 \geq 0, x_2 \geq 0\)). - The **intersection** of these regions is the feasible set. ### **(You would draw this region on graph paper or using a plotting tool.)** --- ## **Step 2: Unconstrained Minimum of \(f(x_1, x_2)\) (part b)** Given: \[ f(x_1, x_2) = \frac{1}{2}x_1^2 + \frac{1}{2}x_2^2 - x_1 - 2x_2 \] ### **Find the unconstrained minimum:** 1. **Set the gradient to zero:** \[ \frac{\partial f}{\partial x_1} = x_1 - 1 = 0 \implies x_1 = 1 \] \[ \frac{\partial f}{\partial x_2} = x_2 - 2 = 0 \implies x_2 = 2 \] **So the unconstrained minimum is at \((x_1, x_2) = (1, 2)\).** 2. **Check if this point is in the feasible set:** - \(2(1) + 3(2) = 2 + 6 = 8 > 6\) (not satisfied) - \(1 + 4(2) = 1 + 8 = 9 > 5\) (not satisfied) - \(x_1 = 1 > 0\), \(x_2 = 2 > 0\) (these are satisfied) **Conclusion:** The unconstrained minimum is **not** in the feasible set. --- ## **Final Answer for (a) and (b):** ### **(a)** - The feasible region is the intersection of the half-planes defined by \(2x_1 + 3x_2 \leq 6\), \(x_1 + 4x_2 \leq 5\), \(x_1 \geq 0\), \(x_2 \geq 0\) in the first quadrant. ### **(b)** - The unconstrained minimum is at \((x_1, x_2) = (1, 2)\), but this point **does not satisfy** the constraints and therefore lies **outside** the feasible set. --- **Let me know if you'd like to proceed with steps (c) and (d) next!**

Let's proceed step by step to address parts (c) and (d), building on what we've already established. --- ### **Part (c): Find the minimum of \(f(x_1, x_2)\) subject to the constraint \(2x_1 + 3x_2 = 6\)** (equality constraint). Is the minimum within the feasible set? #### **Step 1: Set up the constrained minimization problem** - Objective function: \[ f(x_1, x_2) = \frac{1}{2}x_1^2 + \frac{1}{2}x_2^2 - x_1 - 2x_2 \] - Constraint: \[ 2x_1 + 3x_2 = 6 \] --- #### **Step 2: Use the method of Lagrange multipliers** Define the Lagrangian: \[ \mathcal{L} = \frac{1}{2}x_1^2 + \frac{1}{2}x_2^2 - x_1 - 2x_2 + \lambda (2x_1 + 3x_2 - 6) \] Calculate partial derivatives and set to zero: \[ \frac{\partial \mathcal{L}}{\partial x_1} = x_1 - 1 + 2\lambda = 0 \quad \Rightarrow \quad x_1 = 1 - 2\lambda \] \[ \frac{\partial \mathcal{L}}{\partial x_2} = x_2 - 2 + 3\lambda = 0 \quad \Rightarrow \quad x_2 = 2 - 3\lambda \] \[ \frac{\partial \mathcal{L}}{\partial \lambda} = 2x_1 + 3x_2 - 6 = 0 \] Substitute \(x_1, x_2\): \[ 2(1 - 2\lambda) + 3(2 - 3\lambda) = 6 \] \[ 2 - 4\lambda + 6 - 9\lambda = 6 \] \[ (2 + 6) - (4\lambda + 9\lambda) = 6 \] \[ 8 - 13\lambda = 6 \] \[ 13\lambda = 8 - 6 = 2 \] \[ \lambda = \frac{2}{13} \] Now find \(x_1, x_2\): \[ x_1 = 1 - 2 \times \frac{2}{13} = 1 - \frac{4}{13} = \frac{13}{13} - \frac{4}{13} = \frac{9}{13} \] \[ x_2 = 2 - 3 \times \frac{2}{13} = 2 - \frac{6}{13} = \frac{26}{13} - \frac{6}{13} = \frac{20}{13} \] --- #### **Step 3: Check if the solution is within the feasible set** - Since the constraint is an **equality**, the point \((x_1, x_2) = \left(\frac{9}{13}, \frac{20}{13}\right)\) **lies on** the boundary. - Verify constraints (they are satisfied by construction): \[ 2 \times \frac{9}{13} + 3 \times \frac{20}{13} = \frac{18}{13} + \frac{60}{13} = \frac{78}{13} = 6 \] **Yes, it satisfies the equality constraint.** - Is this point **within the feasible set**? Since the original constraints are inequalities, and this point satisfies the equality, it **lies on the boundary** and is therefore **within** the feasible set. --- ### **Final answer for (c):** - The minimum of \(f\) subject to \(2x_1 + 3x_2 = 6\) is at: \[ \boxed{ x_1 = \frac{9}{13}, \quad x_2 = \frac{20}{13} } \] - **Yes**, this point **lies within** the feasible set (on the boundary). --- ### **Part (d): Find the minimum of \(f(x_1, x_2)\) subject to \(x_1 + 4x_2 = 5\). Is the minimum within the feasible set?** #### **Step 1: Set up the problem** - Objective: same \(f\) as above. - Constraint: \[ x_1 + 4x_2 = 5 \] --- #### **Step 2: Use Lagrange multipliers** Define the Lagrangian: \[ \mathcal{L} = \frac{1}{2}x_1^2 + \frac{1}{2}x_2^2 - x_1 - 2x_2 + \mu (x_1 + 4x_2 - 5) \] Calculate derivatives: \[ \frac{\partial \mathcal{L}}{\partial x_1} = x_1 - 1 + \mu = 0 \quad \Rightarrow \quad x_1 = 1 - \mu \] \[ \frac{\partial \mathcal{L}}{\partial x_2} = x_2 - 2 + 4\mu = 0 \quad \Rightarrow \quad x_2 = 2 - 4\mu \] Constraint: \[ x_1 + 4x_2 = 5 \] Substitute: \[ (1 - \mu) + 4(2 - 4\mu) = 5 \] \[ 1 - \mu + 8 - 16\mu = 5 \] \[ (1 + 8) - (\mu + 16 \mu) = 5 \] \[ 9 - 17 \mu = 5 \] \[ 17 \mu = 9 - 5 = 4 \] \[ \mu = \frac{4}{17} \] Calculate \(x_1, x_2\): \[ x_1 = 1 - \frac{4}{17} = \frac{17}{17} - \frac{4}{17} = \frac{13}{17} \] \[ x_2 = 2 - 4 \times \frac{4}{17} = 2 - \frac{16}{17} = \frac{34}{17} - \frac{16}{17} = \frac{18}{17} \] --- #### **Step 3: Check if the solution is within the feasible set** - The point \(\left(\frac{13}{17}, \frac{18}{17}\right)\) **lies on** the boundary \(x_1 + 4x_2 = 5\), satisfying the equality. - Since the constraints are inequalities, the point is **within** the feasible set (on the boundary). --- ### **Final answer for (d):** \[ x_1 = \frac{13}{17}, \quad x_2 = \frac{18}{17} \] - This point **satisfies** the constraint and is **within** the feasible set. --- **Summary:** - **(c)** The constrained minimum on \(2x_1 + 3x_2 = 6\) occurs at \(\left(\frac{9}{13}, \frac{20}{13}\right)\), within the feasible set. - **(d)** The constrained minimum on \(x_1 + 4x_2 = 5\) occurs at \(\left(\frac{13}{17}, \frac{18}{17}\right)\), within the feasible set. Next, you can proceed to part (e), combining these results to find the overall constrained minimum and verify Kuhn-Tucker conditions. Would you like me to help with that?