Let's solve the given optimization problem **using the KKT (Karush-Kuhn-Tucker) conditions** step by step. --- ### **Step 1: Write the Problem** Minimize: \[ f(x_1, x_2) = x_1^2 + x_2^2 \] Subject to: \[ g_1(x_1, x_2) = x_1^2 + 2x_1x_2 + x_2^2 - 1 = 0 \quad \text{(equality constraint)} \] \[ g_2(x_1, x_2) = x_1 - x_2 \leq 0 \quad \text{(inequality constraint)} \] --- ### **Step 2: Write the Lagrangian** Introduce Lagrange multipliers: - \(\lambda\) for the equality constraint \(g_1\) - \(\mu \ge 0\) for the inequality constraint \(g_2\) \[ L(x_1, x_2, \lambda, \mu) = x_1^2 + x_2^2 + \lambda (x_1^2 + 2x_1 x_2 + x_2^2 - 1) + \mu(x_1 - x_2) \] --- ### **Step 3: KKT Conditions** 1. **Stationarity**: \(\nabla_{x_1, x_2} L = 0\) 2. **Primal feasibility**: \(g_1(x_1, x_2) = 0\), \(g_2(x_1, x_2) \leq 0\) 3. **Dual feasibility**: \(\mu \geq 0\) 4. **Complementary slackness**: \(\mu \cdot (x_1 - x_2) = 0\) Let's compute the gradients: #### **Stationarity** \[ \frac{\partial L}{\partial x_1} = 2x_1 + \lambda(2x_1 + 2x_2) + \mu \] \[ \frac{\partial L}{\partial x_2} = 2x_2 + \lambda(2x_1 + 2x_2) - \mu \] Set to zero: \[ 2x_1 + \lambda(2x_1 + 2x_2) + \mu = 0 \tag{1} \] \[ 2x_2 + \lambda(2x_1 + 2x_2) - \mu = 0 \tag{2} \] Also include the primal feasibility constraints: \[ x_1^2 + 2x_1x_2 + x_2^2 = 1 \tag{3} \] \[ x_1 - x_2 \leq 0 \tag{4} \] Dual feasibility: \(\mu \geq 0\) Complementary slackness: \(\mu(x_1 - x_2) = 0 \tag{5}\) --- ## **Step 4: Solve the KKT System** #### **Case 1**: \(x_1 - x_2 < 0 \implies \mu = 0\) Plug \(\mu = 0\) into equations (1) and (2): (1) \(2x_1 + \lambda(2x_1 + 2x_2) = 0\) (2) \(2x_2 + \lambda(2x_1 + 2x_2) = 0\) Subtract (1) from (2): \[ 2x_2 - 2x_1 = 0 \implies x_1 = x_2 \] But this would mean \(x_1 - x_2 = 0\), so actually not possible in this case. So this case yields no solution. --- #### **Case 2**: \(x_1 - x_2 = 0 \implies x_1 = x_2\) By complementary slackness, \(\mu\) can be anything \(\geq 0\). Plug \(x_1 = x_2 = a\) into the constraints: From (3): \(a^2 + 2a^2 + a^2 = 4a^2 = 1 \implies a = \pm \frac{1}{2}\) So \(x_1 = x_2 = \pm \frac{1}{2}\). Now, plug into (1) and (2) to determine \(\lambda, \mu\): \[ 2a + \lambda(4a) + \mu = 0 \] \[ 2a + \lambda(4a) - \mu = 0 \] Subtract: \[ (2a + \lambda(4a) + \mu) - (2a + \lambda(4a) - \mu) = 2\mu = 0 \implies \mu = 0 \] So, \(\mu = 0\). Now, \(2a + \lambda 4a = 0\) If \(a \neq 0\), \(2 + 4\lambda = 0 \implies \lambda = -\frac{1}{2}\) Therefore, the candidate points are: \[ \boxed{ (x_1, x_2) = \left(\frac{1}{2}, \frac{1}{2}\right), \left(-\frac{1}{2}, -\frac{1}{2}\right) } \] Check \(x_1 - x_2 \leq 0\): for both, it's zero, so valid. --- ### **Step 5: Final Candidate Solutions** #### **Final Answer:** The candidate solutions using KKT are: \[ \boxed{ (x_1, x_2) = \left(\frac{1}{2}, \frac{1}{2}\right) \quad \text{and} \quad \left(-\frac{1}{2}, -\frac{1}{2}\right) } \] with corresponding multipliers: - \(\lambda = -\frac{1}{2}\) - \(\mu = 0\) --- **Summary of Steps:** 1. Write the Lagrangian. 2. Write the KKT conditions. 3. Solve stationarity and constraints. 4. Use complementary slackness to separate cases. 5. Find candidate points that satisfy all KKT conditions.

Let's carefully solve the given problem step-by-step, providing detailed calculations and reasoning. --- ## **Given Problem** Minimize: \[ f(x_1, x_2) = x_1^2 + x_2^2 \] Subject to the constraints: \[ \begin{cases} g_1(x_1, x_2) = x_1^2 + 2x_1 x_2 + x_2^2 - 1 = 0 \quad \text{(equality constraint)} \\ g_2(x_1, x_2) = x_1 - x_2 \leq 0 \quad \text{(inequality constraint)} \end{cases} \] --- ## **Step 1: Formulate the Lagrangian** Introduce multipliers: - \(\lambda\) for the equality constraint \(g_1\) - \(\mu \ge 0\) for the inequality constraint \(g_2\) The Lagrangian: \[ \boxed{ L(x_1, x_2, \lambda, \mu) = x_1^2 + x_2^2 + \lambda (x_1^2 + 2x_1 x_2 + x_2^2 - 1) + \mu (x_1 - x_2) } \] --- ## **Step 2: State the KKT Conditions** The conditions are: 1. **Stationarity**: \[ \nabla_{x} L = 0 \] which gives: \[ \frac{\partial L}{\partial x_1} = 0, \quad \frac{\partial L}{\partial x_2} = 0 \] 2. **Primal feasibility**: \[ g_1(x_1, x_2) = 0, \quad g_2(x_1, x_2) \le 0 \] 3. **Dual feasibility**: \[ \mu \ge 0 \] 4. **Complementary slackness**: \[ \mu \cdot (x_1 - x_2) = 0 \] --- ## **Step 3: Compute the gradients** ### Partial derivatives: \[ \frac{\partial L}{\partial x_1} = 2x_1 + \lambda (2x_1 + 2x_2) + \mu \] \[ \frac{\partial L}{\partial x_2} = 2x_2 + \lambda (2x_1 + 2x_2) - \mu \] Set to zero: \[ \boxed{ \begin{cases} 2x_1 + 2\lambda (x_1 + x_2) + \mu = 0 \quad (1) \\ 2x_2 + 2\lambda (x_1 + x_2) - \mu = 0 \quad (2) \end{cases} } \] --- ## **Step 4: Solve the KKT system** ### Combine equations (1) and (2): Subtract (2) from (1): \[ (2x_1 - 2x_2) + 2\lambda (x_1 + x_2 - x_1 - x_2) + (\mu + \mu) = 0 \] Simplify: \[ 2(x_1 - x_2) + 2\mu = 0 \] which yields: \[ \boxed{ x_1 - x_2 = -\mu } \] --- ### **Case 1: \(\mu = 0\)** If \(\mu = 0\), then: \[ x_1 - x_2 = 0 \implies x_1 = x_2 \] Plug into the equality constraint \(g_1\): \[ x_1^2 + 2x_1 x_2 + x_2^2 = 1 \] Since \(x_1 = x_2 = a\): \[ a^2 + 2a^2 + a^2 = 4a^2 = 1 \implies a = \pm \frac{1}{2} \] Thus, candidate points: \[ \boxed{ \left( \frac{1}{2}, \frac{1}{2} \right), \quad \left( -\frac{1}{2}, -\frac{1}{2} \right) } \] Now, determine the multipliers \(\lambda\): From (1): \[ 2a + 2\lambda (a + a) + 0 = 0 \] \[ 2a + 4a \lambda = 0 \] Solve for \(\lambda\): \[ 4a \lambda = -2a \implies \lambda = -\frac{2a}{4a} = -\frac{1}{2} \] Since \(a \neq 0\), the above holds for both points: - For \(a = \frac{1}{2}\): \[ \lambda = -\frac{1}{2} \] - For \(a = -\frac{1}{2}\): \[ \lambda = -\frac{1}{2} \] Check the **inequality constraint**: \[ x_1 - x_2 \le 0 \] - At both points, \(x_1 = x_2\), so the constraint is active (\(x_1 - x_2 = 0\)). - Complementary slackness: \(\mu \cdot 0 = 0 \Rightarrow \text{satisfied}\). - Since \(\mu = 0 \ge 0\), the dual feasibility holds. --- ### **Case 2: \(\mu > 0\)** Recall: \[ x_1 - x_2 = -\mu \] From the stationarity equations (1) and (2): \[ 2x_1 + 2\lambda (x_1 + x_2) + \mu = 0 \] \[ 2x_2 + 2\lambda (x_1 + x_2) - \mu = 0 \] Add these two equations: \[ (2x_1 + 2x_2) + 4\lambda (x_1 + x_2) = 0 \] Define \(S = x_1 + x_2\). Then: \[ 2(x_1 + x_2) + 4\lambda S = 0 \implies 2S + 4\lambda S = 0 \] \[ S (2 + 4\lambda) = 0 \] Two possibilities: 1. \(S = 0\) 2. \(2 + 4\lambda = 0 \Rightarrow \lambda = -\frac{1}{2}\) --- ### **Subcase 2.1: \(S = 0\)** From \(S = x_1 + x_2 = 0\), hence: \[ x_2 = -x_1 \] From the earlier relation: \[ x_1 - x_2 = -\mu \] But: \[ x_2 = -x_1 \Rightarrow x_1 - (-x_1) = 2x_1 = -\mu \] \[ \Rightarrow x_1 = -\frac{\mu}{2} \] \[ x_2 = -x_1 = \frac{\mu}{2} \] Now, from the equality constraint \(g_1\): \[ x_1^2 + 2x_1 x_2 + x_2^2 = 1 \] Plugging in \(x_2 = -x_1\): \[ x_1^2 + 2x_1(-x_1) + (-x_1)^2 = 1 \] \[ x_1^2 - 2x_1^2 + x_1^2 = 1 \] \[ (1 - 2 + 1) x_1^2 = 1 \Rightarrow 0 \cdot x_1^2 = 1 \] Contradiction! No solution here. --- ### **Subcase 2.2: \(\lambda = -\frac{1}{2}\)** Recall from the stationarity equations: Equation (1): \[ 2x_1 + 2\lambda (x_1 + x_2) + \mu = 0 \] \[ 2x_1 + 2 \left(-\frac{1}{2}\right) (x_1 + x_2) + \mu = 0 \] \[ 2x_1 - (x_1 + x_2) + \mu = 0 \] \[ (2x_1 - x_1 - x_2) + \mu = 0 \Rightarrow (x_1 - x_2) + \mu = 0 \] But earlier, we have: \[ x_1 - x_2 = -\mu \] Substitute: \[ (-\mu) + \mu = 0 \Rightarrow 0=0 \] This is consistent. Now, from the previous relation: \[ x_1 - x_2 = -\mu \] and the equality constraint: \[ x_1^2 + 2x_1 x_2 + x_2^2 = 1 \] Express \(x_2\) in terms of \(x_1\): \[ x_2 = x_1 + \mu \] Plug into the constraint: \[ x_1^2 + 2x_1 (x_1 + \mu) + (x_1 + \mu)^2 = 1 \] Expanding: \[ x_1^2 + 2x_1^2 + 2x_1 \mu + x_1^2 + 2x_1 \mu + \mu^2 = 1 \] Combine like terms: \[ ( x_1^2 + 2x_1^2 + x_1^2 ) + ( 2x_1 \mu + 2x_1 \mu ) + \mu^2 = 1 \] \[ 4x_1^2 + 4x_1 \mu + \mu^2 = 1 \] Recall that \(x_2 = x_1 + \mu\) and \(x_1 - x_2 = -\mu\). Now, the key is to find \(x_1, \mu\) satisfying this quadratic: \[ 4x_1^2 + 4x_1 \mu + \mu^2 = 1 \] Note that \(\mu \ge 0\), and from earlier, the relation \(x_2 = x_1 + \mu\). --- ## **Step 5: Find explicit solutions** Suppose for simplicity that \(\mu\) is known, then: \[ 4x_1^2 + 4x_1 \mu + \mu^2 = 1 \] Treat this as quadratic in \(x_1\): \[ 4x_1^2 + 4 \mu x_1 + (\mu^2 - 1) = 0 \] Discriminant: \[ \Delta = (4\mu)^2 - 4 \times 4 \times (\mu^2 - 1) = 16 \mu^2 - 16 (\mu^2 - 1) = 16 \mu^2 - 16 \mu^2 + 16 = 16 \] Since \(\Delta = 16\), real solutions exist: \[ x_1 = \frac{-4 \mu \pm \sqrt{16}}{2 \times 4} = \frac{-4 \mu \pm 4}{8} = \frac{- \mu \pm 1}{2} \] Hence: \[ x_1 = \frac{-\mu + 1}{2} \quad \text{or} \quad x_1 = \frac{-\mu - 1}{2} \] Corresponding \(x_2\): \[ x_2 = x_1 + \mu \] ### **Case 2.2a:** \[ x_1 = \frac{-\mu + 1}{2} \] \[ x_2 = x_1 + \mu = \frac{-\mu + 1}{2} + \mu = \frac{-\mu + 1 + 2 \mu}{2} = \frac{\mu + 1}{2} \] Since \(x_1 - x_2 = -\mu\): \[ x_1 - x_2 = \frac{-\mu + 1}{2} - \frac{\mu + 1}{2} = \frac{-\mu + 1 - \mu - 1}{2} = \frac{-2\mu}{2} = -\mu \] which is consistent. Now, **\(\mu \ge 0\)**, so: \[ \boxed{ x_1 = \frac{1 - \mu}{2}, \quad x_2 = \frac{1 + \mu}{2} } \] Check the inequality constraint \(x_1 - x_2 \le 0\): \[ x_1 - x_2 = -\mu \le 0 \] which holds for \(\mu \ge 0\). --- ## **Step 6: Summary of candidate solutions** From the above: - **Candidates with \(\mu = 0\)**

The candidate solutions are \(\boxed{\left(\frac{1}{2}, \frac{1}{2}\right)}\) and \(\boxed{\left(-\frac{1}{2}, -\frac{1}{2}\right)}\).