Let's solve the problem in **2 steps** as requested, with explanations at each step: --- ### **Step 1: Feasibility of the Point \((\frac{9}{5}, \frac{6}{5})\)** #### **Check Constraints:** Let \( x_1 = \frac{9}{5} \), \( x_2 = \frac{6}{5} \). **Constraint 1:** \( x_1^2 + x_2^2 \leq 5 \) \[ \left(\frac{9}{5}\right)^2 + \left(\frac{6}{5}\right)^2 = \frac{81}{25} + \frac{36}{25} = \frac{117}{25} = 4.68 \leq 5 \] **Constraint 2:** \( x_1 + x_2 \leq 3 \) \[ \frac{9}{5} + \frac{6}{5} = \frac{15}{5} = 3 \leq 3 \] **Constraint 3:** \( x_1, x_2 \geq 0 \) \[ x_1 = \frac{9}{5} \geq 0, \quad x_2 = \frac{6}{5} \geq 0 \] **Conclusion:** The point \(\left(\frac{9}{5}, \frac{6}{5}\right)\) **satisfies all constraints** and is **feasible**. --- ### **Step 2: Verifying Local Optimality and Uniqueness** #### **Objective Function:** \[ f(x_1, x_2) = (x_1 - 3)^2 + (x_2 - 2)^2 \] This is a convex quadratic function (sum of squares), minimized at \( (3, 2) \). But \( (3, 2) \) is **outside** the feasible region (since \( 3^2 + 2^2 = 13 > 5 \)). #### **Is \(\left(\frac{9}{5}, \frac{6}{5}\right)\) the Local Minimum?** To check, let's see if it's the **projection** of \((3, 2)\) onto the feasible set. The constraints \( x_1^2 + x_2^2 \leq 5 \) and \( x_1 + x_2 \leq 3 \) are both active at this point. Let's check if the point lies on the boundary: - \( x_1^2 + x_2^2 = 4.68 < 5 \) → Not tight (but close). - \( x_1 + x_2 = 3 \) → Tight. The closest point to \((3, 2)\) on the feasible set will be on the boundary, likely where \( x_1 + x_2 = 3 \). Let's minimize \( (x_1 - 3)^2 + (x_2 - 2)^2 \) subject to \( x_1 + x_2 = 3 \) and \( x_1^2 + x_2^2 \leq 5 \). **Using Lagrange Multipliers:** Set up for \( x_1 + x_2 = 3 \): Let \( x_2 = 3 - x_1 \), substitute into the constraint: \[ x_1^2 + (3 - x_1)^2 \leq 5 \\ x_1^2 + 9 - 6x_1 + x_1^2 \leq 5 \\ 2x_1^2 - 6x_1 + 9 \leq 5 \\ 2x_1^2 - 6x_1 + 4 \leq 0 \] This quadratic has roots: \[ 2x_1^2 - 6x_1 + 4 = 0 \\ x_1 = \frac{6 \pm \sqrt{36 - 32}}{4} = \frac{6 \pm 2}{4} = 2, 1 \] So \( 1 \leq x_1 \leq 2 \), \( x_2 = 3 - x_1 \). **Now, minimize \( (x_1 - 3)^2 + (x_2 - 2)^2 \):** \[ = (x_1 - 3)^2 + (3 - x_1 - 2)^2 = (x_1 - 3)^2 + (1 - x_1)^2 \] \[ = (x_1^2 - 6x_1 + 9) + (1 - 2x_1 + x_1^2) \] \[ = 2x_1^2 - 8x_1 + 10 \] Take derivative and set to zero: \[ \frac{d}{dx_1} [2x_1^2 - 8x_1 + 10] = 4x_1 - 8 = 0 \implies x_1 = 2 \] Check if \( x_1 = 2 \) is in feasible interval [1,2]: Yes. So \( x_1^* = 2 \), \( x_2^* = 1 \). Check if this satisfies \( x_1^2 + x_2^2 \leq 5 \): \[ 2^2 + 1^2 = 4 + 1 = 5 \leq 5 \] Yes, it is on the boundary. #### **Final Answers** #### **Explanation:** - The point \( \left(\frac{9}{5}, \frac{6}{5}\right) \) is **feasible** but **not the local minimum**; the optimal solution is at \( (2, 1) \). - The solution is **unique** because the objective function is strictly convex and the feasible region is convex; thus, the minimum is unique. --- ## **Final Answers** - **The point \( \left(\frac{9}{5}, \frac{6}{5}\right) \) is not a local minimum.** - **The unique optimal solution is \( (x_1^*, x_2^*) = (2, 1) \).** - **The optimal solution is unique because the problem is strictly convex.**

Sure! Here's a clear, human-like explanation: First, I checked whether the point (9/5, 6/5) is feasible. It turns out it satisfies all the constraints, so it's feasible. But just because it's feasible doesn't mean it's the best (or optimal) point. Next, I looked at where the actual minimum might be. The objective function we're trying to minimize is a sum of squares, which is convex and would normally be minimized at (3, 2). However, (3, 2) isn't inside our feasible region because it violates the constraints. So, I examined the boundary of the feasible region, especially where the constraints are tight, to find the point closest to (3, 2). Doing that, I found that the best point on the boundary is (2, 1), which satisfies all the constraints perfectly and gives the smallest value of the objective function. Since the problem involves a convex function over a convex region, the optimal solution is unique. Therefore, (2, 1) is the actual minimal point, and the point (9/5, 6/5) is not a local minimum.