Let's solve the problem step-by-step: ### **Given Problem** \[ \text{(P)} \quad \min \quad (x_1 - 3)^2 + (x_2 - 2)^2 \] subject to: \[ \begin{align*} x_1^2 + x_2^2 &\leq 5 \\ x_1 + x_2 &\leq 4 \\ x_1 - x_2 &\geq 0 \end{align*} \] --- ### **1. Find the optimal solution of P** #### **Step 1: Understanding the Objective Function** The objective function is the squared distance from the point \((3, 2)\). We want to find the point within the feasible region that is closest to \((3, 2)\). #### **Step 2: Feasible Region** - \(x_1^2 + x_2^2 \leq 5\) — interior of a circle of radius \(\sqrt{5}\) centered at (0,0) - \(x_1 + x_2 \leq 4\) — below the line \(x_1 + x_2 = 4\) - \(x_1 - x_2 \geq 0\) — above the line \(x_1 = x_2\) (or equivalently, \(x_1 \geq x_2\)) #### **Step 3: Check if (3, 2) is Feasible** - \(3^2 + 2^2 = 9 + 4 = 13 > 5\): **Not feasible** - \(3 + 2 = 5 > 4\): **Not feasible** - \(3 - 2 = 1 \geq 0\): **OK** So, the unconstrained minimizer is not feasible. The solution must be on the boundary. #### **Step 4: Solve using Lagrange Multipliers (KKT conditions)** Let’s denote: - \(g_1(x_1, x_2) = x_1^2 + x_2^2 - 5 \leq 0\) - \(g_2(x_1, x_2) = x_1 + x_2 - 4 \leq 0\) - \(g_3(x_1, x_2) = x_2 - x_1 \leq 0\) Let’s check intersection points of boundaries. --- #### **Step 5: Try Intersections** Let's try the intersection of the circle and the line \(x_1 + x_2 = 4\): Let \(x_1 + x_2 = 4\). Substitute \(x_2 = 4 - x_1\) into the circle: \[ x_1^2 + (4 - x_1)^2 = 5 \\ x_1^2 + 16 - 8x_1 + x_1^2 = 5 \\ 2x_1^2 - 8x_1 + 16 = 5 \\ 2x_1^2 - 8x_1 + 11 = 0 \\ x_1^2 - 4x_1 + 5.5 = 0 \] Solve the quadratic: \[ x_1 = \frac{4 \pm \sqrt{16 - 4 \times 1 \times 5.5}}{2} = \frac{4 \pm \sqrt{16 - 22}}{2} = \frac{4 \pm \sqrt{-6}}{2} \] This has **no real solution**, so the circle and the line do not intersect within the feasible region. --- #### **Step 6: Try Intersection of \(x_1^2 + x_2^2 = 5\) and \(x_1 = x_2\)** Set \(x_1 = x_2 = t\): \[ t^2 + t^2 = 5 \implies t^2 = 2.5 \implies t = \pm \sqrt{2.5} \approx \pm 1.58 \] Now, check constraints: - \(x_1 + x_2 = 2t \leq 4 \implies t \leq 2\) - For \(t = 1.58\), \(2 \times 1.58 = 3.16 < 4\), so feasible. So the point \((1.58, 1.58)\) is feasible. Calculate objective at \((1.58, 1.58)\): \[ (x_1 - 3)^2 + (x_2 - 2)^2 = (1.58 - 3)^2 + (1.58 - 2)^2 = (-1.42)^2 + (-0.42)^2 = 2.0164 + 0.1764 = 2.1928 \] --- #### **Step 7: Check Points on the Circle Boundary** Try maximizing \(x_1\) (to get closer to 3), with \(x_1^2 + x_2^2 = 5\), \(x_1 \geq x_2\): Let \(x_2 = 0\): \[ x_1^2 = 5 \implies x_1 = \sqrt{5} \approx 2.236 \] Check constraints: - \(x_1 + x_2 = 2.236 \leq 4\) ✓ - \(x_1 \geq x_2\) ✓ Point: \((2.236, 0)\) Objective: \[ (2.236 - 3)^2 + (0 - 2)^2 = (-0.764)^2 + (-2)^2 = 0.584 + 4 = 4.584 \] Try \(x_2 = 0\), \(x_1 = 2\): - \(x_1^2 + x_2^2 = 4\) < 5 (inside the circle) - \(x_1 + x_2 = 2\) < 4 But objective is worse than \((1.58, 1.58)\). --- #### **Step 8: Try on Boundary \(x_1 + x_2 = 4\), \(x_1 \geq x_2\)** Let \(x_1 = 4 - x_2\), \(x_1 \geq x_2 \implies 4 - x_2 \geq x_2 \implies x_2 \leq 2\) Plug into circle: \[ (4 - x_2)^2 + x_2^2 = 5 \\ 16 - 8x_2 + x_2^2 + x_2^2 = 5 \\ 16 - 8x_2 + 2x_2^2 = 5 \\ 2x_2^2 - 8x_2 + 11 = 0 \\ x_2^2 - 4x_2 + 5.5 = 0 \] As before, discriminant is negative, so no real solution. --- #### **Step 9: Try on \(x_1^2 + x_2^2 = 5\), maximizing \(x_1\), \(x_1 \geq x_2\), \(x_1 + x_2 \leq 4\)** Try \(x_2 = 1\): \[ x_1^2 + 1^2 = 5 \implies x_1^2 = 4 \implies x_1 = 2 \] Check \(x_1 + x_2 = 2 + 1 = 3 \leq 4\) Check \(x_1 \geq x_2\): \(2 \geq 1\) Objective: \[ (2-3)^2 + (1-2)^2 = (-1)^2 + (-1)^2 = 1 + 1 = 2 \] This is better than before. Try \(x_2 = 0.5\): \[ x_1^2 + 0.25 = 5 \implies x_1^2 = 4.75 \implies x_1 \approx 2.179 x_1 + x_2 \approx 2.179 + 0.5 = 2.679 \leq 4 x_1 \geq x_2 \] Objective: \[ (2.179-3)^2 + (0.5-2)^2 = (-0.821)^2 + (-1.5)^2 = 0.674 + 2.25 = 2.924 \] Worse than (2,1). Try \(x_2 = 1.5\): \[ x_1^2 + 2.25 = 5 \implies x_1^2 = 2.75 \implies x_1 \approx 1.658 x_1 + x_2 \approx 1.658 + 1.5 = 3.158 \leq 4 x_1 \geq x_2? 1.658 < 1.5 \text{ (not valid)} \] So, not valid. --- #### **Step 10: Try \(x_2 = -1\):\** \[ x_1^2 + 1 = 5 \implies x_1^2 = 4 \implies x_1 = 2 x_1 + x_2 = 2 + (-1) = 1 \leq 4 x_1 \geq x_2: 2 \geq -1 \] Objective: \[ (2-3)^2 + (-1-2)^2 = 1 + 9 = 10 \] Not better. --- #### **Step 11: Try \(x_2 = 0\):\** \[ x_1^2 = 5 \implies x_1 = \sqrt{5} = 2.236 x_1 + x_2 = 2.236 \leq 4 x_1 \geq x_2: 2.236 \geq 0 \] Objective: \[ (2.236-3)^2 + (0-2)^2 = (-0.764)^2 + (-2)^2 = 0.584 + 4 = 4.584 \] Not better. --- #### **Step 12: Try \(x_2 = 1.2\):\** \[ x_1^2 + (1.2)^2 = 5 \implies x_1^2 = 5 - 1.44 = 3.56 \implies x_1 = 1.888 x_1 + x_2 = 1.888 + 1.2 = 3.088 \leq 4 x_1 \geq x_2: 1.888 \geq 1.2 \] Objective: \[ (1.888-3)^2 + (1.2-2)^2 = (-1.112)^2 + (-0.8)^2 = 1.236 + 0.64 = 1.876 \] This is even better. Try \(x_2 = 1.5\): \[ x_1^2 = 5 - 2.25 = 2.75 \implies x_1 = 1.658 x_1 + x_2 = 1.658 + 1.5 = 3.158 \leq 4 x_1 \geq x_2: 1.658 \geq 1.5 \] Objective: \[ (1.658-3)^2 + (1.5-2)^2 = (-1.342)^2 + (-0.5)^2 = 1.801 + 0.25 = 2.051 \] A bit worse, so the best so far is at \(x_2 = 1.2, x_1 = 1.888\). --- #### **Step 13: Try \(x_2 = 1.4\):\** \[ x_1^2 = 5 - 1.96 = 3.04 \implies x_1 = 1.744 x_1 + x_2 = 1.744 + 1.4 = 3.144 \leq 4 x_1 \geq x_2: 1.744 \geq 1.4 \] Objective: \[ (1.744-3)^2 + (1.4-2)^2 = (-1.256)^2 + (-0.6)^2 = 1.578 + 0.36 = 1.938 \] Worse than at \(x_2 = 1.2\). --- #### **Step 14: Try \(x_2 = 1.1\):\** \[ x_1^2 = 5 - 1.21 = 3.79 \implies x_1 = 1.947 x_1 + x_2 = 1.947 + 1.1 = 3.047 \leq 4 x_1 \geq x_2 \] Objective: \[ (1.947-3)^2 + (1.1-2)^2 = (-1.053)^2 + (-0.9)^2 = 1.109 + 0.81 = 1.919 \] Best so far is at \(x_2 =

Let's analyze the problem carefully and proceed step-by-step with detailed calculations. --- ### **Given problem:** Minimize the distance squared from \((3,2)\): \[ \boxed{ \text{Minimize } f(x_1, x_2) = (x_1 - 3)^2 + (x_2 - 2)^2 } \] subject to the constraints: \[ \begin{cases} x_1^2 + x_2^2 \leq 5 \quad &\text{(circle of radius } \sqrt{5} \text{)} \\ x_1 + x_2 \leq 4 \quad &\text{(line)} \\ x_1 - x_2 \geq 0 \quad &\text{(region above } x_1 = x_2 \text{)} \end{cases} \] --- ### **Step 1: Understanding the problem** - The point \((3,2)\) is outside the feasible region because: \[ 3^2 + 2^2 = 9 + 4 = 13 > 5, \] so not inside the circle. - The closest feasible point to \((3,2)\) will be on the boundary of the feasible region (since the objective is a quadratic with a minimum). --- ### **Step 2: Feasible region boundaries** - **Circle boundary:** \[ x_1^2 + x_2^2 = 5 \] - **Line boundary:** \[ x_1 + x_2 = 4 \] - **Line \(x_1 = x_2\):** \[ x_1 - x_2 = 0 \] --- ### **Step 3: Candidate points** Potential solutions are intersections of the boundary lines: - Circle and the line \(x_1 + x_2 = 4\), - Circle and the line \(x_1 = x_2\), - Points on the boundary that satisfy the inequalities. --- ### **Step 4: Intersection of circle and line \(x_1 + x_2 = 4\)** Substitute \(x_2 = 4 - x_1\) into the circle: \[ x_1^2 + (4 - x_1)^2 = 5 \] \[ x_1^2 + 16 - 8x_1 + x_1^2 = 5 \] \[ 2x_1^2 - 8x_1 + 16 = 5 \] \[ 2x_1^2 - 8x_1 + 11 = 0 \] Divide through by 2: \[ x_1^2 - 4x_1 + 5.5 = 0 \] Discriminant: \[ \Delta = (-4)^2 - 4 \times 1 \times 5.5 = 16 - 22 = -6 < 0 \] **No real solutions**, so the circle and the line \(x_1 + x_2 = 4\) do **not** intersect. --- ### **Step 5: Intersection of circle and \(x_1 = x_2\)** Set \(x_1 = x_2 = t\): \[ 2t^2 = 5 \implies t^2 = 2.5 \implies t = \pm \sqrt{2.5} \approx \pm 1.58 \] Check whether these points satisfy the inequalities: - For \(t = 1.58\): \[ x_1 = x_2 = 1.58 \] Check the constraints: \[ x_1 + x_2 = 3.16 \leq 4 \quad \checkmark \] \[ x_1 - x_2 = 0 \geq 0 \quad \checkmark \] \[ x_1^2 + x_2^2 = 2.5 + 2.5 = 5 \quad \checkmark \] **Feasible point:** \[ \boxed{ (1.58, 1.58) } \] Calculate the distance squared to \((3, 2)\): \[ f = (1.58 - 3)^2 + (1.58 - 2)^2 = (-1.42)^2 + (-0.42)^2 = 2.0164 + 0.1764 = \boxed{2.1928} \] --- ### **Step 6: Check other candidate points** Since the previous candidate is feasible and close, check whether points on the circle closer to \((3,2)\) exist at the boundary. --- ### **Step 7: Points on the circle boundary that satisfy \(x_1 \geq x_2\)** Recall the circle: \[ x_1^2 + x_2^2 = 5 \] - To minimize distance to \((3,2)\), consider parametric form: \[ x_1 = \sqrt{5} \cos \theta,\quad x_2 = \sqrt{5} \sin \theta \] with \(\sqrt{5} \approx 2.236\). --- ### **Step 8: Find the \(\theta\) minimizing the squared distance** Objective: \[ f(\theta) = (\sqrt{5} \cos \theta - 3)^2 + (\sqrt{5} \sin \theta - 2)^2 \] Expand: \[ f(\theta) = (\sqrt{5} \cos \theta)^2 - 2 \times 3 \times \sqrt{5} \cos \theta + 9 + (\sqrt{5} \sin \theta)^2 - 2 \times 2 \times \sqrt{5} \sin \theta + 4 \] Simplify: \[ f(\theta) = 5 \cos^2 \theta - 6 \sqrt{5} \cos \theta + 9 + 5 \sin^2 \theta - 4 \sqrt{5} \sin \theta + 4 \] Since \(\cos^2 \theta + \sin^2 \theta = 1\): \[ f(\theta) = 5 (1) - 6 \sqrt{5} \cos \theta - 4 \sqrt{5} \sin \theta + (9 + 4) = 5 - 6 \sqrt{5} \cos \theta - 4 \sqrt{5} \sin \theta + 13 \] \[ f(\theta) = 18 - 6 \sqrt{5} \cos \theta - 4 \sqrt{5} \sin \theta \] --- ### **Step 9: Minimize \(f(\theta)\)** The function: \[ f(\theta) = 18 - A \cos \theta - B \sin \theta \] where: \[ A = 6 \sqrt{5} \approx 6 \times 2.236 = 13.416 \] \[ B = 4 \sqrt{5} \approx 4 \times 2.236 = 8.944 \] The minimum of: \[ f(\theta) = 18 - R \cos (\theta - \phi) \] where: \[ R = \sqrt{A^2 + B^2} = \sqrt{(13.416)^2 + (8.944)^2} \] Calculate \(R\): \[ A^2 = 13.416^2 \approx 180.1 \] \[ B^2 = 8.944^2 \approx 80.0 \] \[ R \approx \sqrt{180.1 + 80.0} = \sqrt{260.1} \approx 16.125 \] The minimum occurs when \(\cos (\theta - \phi) = 1\): \[ f_{\text{min}} = 18 - R = 18 - 16.125 = 1.875 \] Corresponds to: \[ f(\theta)_{\text{min}} \approx 1.875 \] Now, find the \(\theta\) that achieves this: \[ \cos (\theta - \phi) = 1 \Rightarrow \theta = \phi \] Where: \[ \phi = \arctan \left( \frac{B}{A} \right) = \arctan \left( \frac{8.944}{13.416} \right) \approx \arctan(0.667) \approx 33.7^\circ \] Corresponding point: \[ x_1 = \sqrt{5} \cos \phi \approx 2.236 \times \cos 33.7^\circ \approx 2.236 \times 0.832 = 1.86 \] \[ x_2 = \sqrt{5} \sin \phi \approx 2.236 \times 0.555 = 1.24 \] Check constraints: - \(x_1 + x_2 \approx 1.86 + 1.24 = 3.10 \leq 4\) ✓ - \(x_1 \geq x_2\): \(1.86 \geq 1.24\) ✓ - \(x_1^2 + x_2^2 \approx 1.86^2 + 1.24^2 = 3.46 + 1.54 = 5\) ✓ **Candidate point: \((1.86, 1.24)\)** Calculate the squared distance: \[ f = (1.86 - 3)^2 + (1.24 - 2)^2 = (-1.14)^2 + (-0.76)^2 = 1.30 + 0.58 = 1.88 \] which is **smaller** than previous candidate \(2.1928\). --- ### **### **Final comparison of candidates:** | Candidate point | Distance squared to \((3,2)\) | Feasible? | Remarks | |------------------|------------------------------|------------|---------| | \((1.58, 1.58)\) | 2.1928 | Yes | From earlier | | \((1.86, 1.24)\) | **1.88** | Yes | From optimization | The **best candidate** is: \[ \boxed{ \boxed{ \textbf{Optimal point } \approx (1.86, 1.24) } } \] with the **minimum squared distance**: \[ \boxed{ \boxed{ \textbf{Minimum value } \approx 1.88 } } \] --- ### **Final Answer:** \[ \boxed{ \boxed{ \textbf{Optimal point } \approx (1.86,\ 1.24), \quad \text{Minimum distance squared} \approx 1.88 } } \] This point minimizes the distance squared from \((3,2)\) within the feasible region defined by the constraints.