(a) \( \operatorname{minimize} x^{2}+(y-1)^{2} \) subject to \( x^{2}+y^{2} \leq 2 \) \( y^{2}-x \leq-2 \) \( -x-y \leq-2 \)

To solve the optimization problem, we need to minimize the objective function \( f(x, y) = x^2 + (y - 1)^2 \) subject to the given constraints: 1. \( x^2 + y^2 \leq 2 \) 2. \( y^2 - x \leq -2 \) (which can be rewritten as \( y^2 \leq x - 2 \)) 3. \( -x - y \leq -2 \) (which can be rewritten as \( x + y \geq 2 \)) ### Step 1: Analyze the Constraints 1. **Circle Constraint**: The first constraint \( x^2 + y^2 \leq 2 \) describes a circle centered at the origin with radius \( \sqrt{2} \). 2. **Parabola Constraint**: The second constraint \( y^2 \leq x - 2 \) describes a region to the left of the parabola that opens to the right and is shifted to the right by 2 units. 3. **Line Constraint**: The third constraint \( x + y \geq 2 \) describes a half-plane above the line \( x + y = 2 \). ### Step 2: Find the Feasible Region To find the feasible region, we need to determine where these constraints intersect. 1. **Circle**: \( x^2 + y^2 = 2 \) 2. **Parabola**: \( y^2 = x - 2 \) 3. **Line**: \( x + y = 2 \) ### Step 3: Find Intersection Points 1. **Intersection of Circle and Line**: Substitute \( y = 2 - x \) into the circle equation: \[ x^2 + (2 - x)^2 = 2 \] Expanding and simplifying: \[ x^2 + (4 - 4x + x^2) = 2 \implies 2x^2 - 4x + 2 = 0 \implies x^2 - 2x + 1 = 0 \implies (x - 1)^2 = 0 \] Thus, \( x = 1 \) and \( y = 1 \). So, the intersection point is \( (1, 1) \). 2. **Intersection of Circle and Parabola**: Substitute \( y^2 = x - 2 \) into the circle equation: \[ x^2 + (x - 2) = 2 \implies x^2 + x - 2 = 2 \implies x^2 + x - 4 = 0 \] Using the quadratic formula: \[ x = \frac{-1 \pm \sqrt{1 + 16}}{2} = \frac{-1 \pm \sqrt{17}}{2} \] Calculate \( y \) for both \( x \) values: - For \( x = \frac{-1 + \sqrt{17}}{2} \): \[ y^2 = \frac{-1 + \sqrt{17}}{2} - 2 = \frac{-5 + \sqrt{17}}{2} \] - For \( x = \frac{-1 - \sqrt{17}}{2} \), this will yield negative values for \( y^2 \), which is not valid. 3. **Intersection of Line and Parabola**: Substitute \( y = 2 - x \) into \( y^2 = x - 2 \): \[ (2 - x)^2 = x - 2 \implies 4 - 4x + x^2 = x - 2 \implies x^2 - 5x + 6 = 0 \] Factoring gives: \[ (x - 2)(x - 3) = 0 \implies x = 2 \text{ or } x = 3 \] - For \( x = 2 \), \( y = 0 \). - For \( x = 3 \), \( y = -1 \). ### Step 4: Evaluate the Objective Function Now we have the points of intersection: 1. \( (1, 1) \) 2. \( (2, 0) \) 3. \( (3, -1) \) (not feasible since it does not satisfy the circle constraint) Evaluate \( f(x, y) \) at the feasible points: 1. \( f(1, 1) = 1^2 + (1 - 1)^2 = 1 \) 2. \( f(2, 0) = 2^2 + (0 - 1)^2 = 4 + 1 = 5 \) ### Conclusion The minimum value of the objective function occurs at the point \( (1, 1) \) with a minimum value of \( 1 \). Thus, the solution to the optimization problem is: \[ \text{Minimum value: } 1 \text{ at } (x, y) = (1, 1). \]