To determine whether the given problem is convex and to find the Lagrange dual, let’s proceed step by step.

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Step 1: Problem Statement

The optimization problem is:

$\text{Minimize: } f(x) = \frac{1}{2}x_1^2 + \frac{1}{2}x_2^2 + 2x_1x_2 + x_1 - x_2 - x_3$

Subject to:

1. $x_1 + x_2 + x_3 \leq 1$
2. $x_3 \leq 1$

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Step 2: Convexity Analysis

Objective Function Analysis

The objective function can be written as: $f(x) = \frac{1}{2}x_1^2 + \frac{1}{2}x_2^2 + 2x_1x_2 + x_1 - x_2 - x_3$

The quadratic terms suggest a second-order polynomial. To check convexity, we evaluate the Hessian of $f(x)$, which is the matrix of second partial derivatives of $f(x)$:

1 & 2 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ The Hessian must be positive semidefinite for $$f(x)$$ to be convex. For $$H$$ to be positive semidefinite, all its eigenvalues must be non-negative. The eigenvalues of $$H$$ are: 1. $$\lambda_1 = 3$$ (positive) 2. $$\lambda_2 = -1$$ (negative) 3. $$\lambda_3 = 0$$ Since one of the eigenvalues ($$\lambda_2 = -1$$) is negative, the Hessian is not positive semidefinite. Therefore, **the objective function is not convex**. #### **Constraint Analysis** The constraints are: 1. $$x_1 + x_2 + x_3 \leq 1$$ 2. $$x_3 \leq 1$$ These constraints are linear and thus convex. However, the non-convexity of the objective function makes the overall optimization problem **non-convex**. --- ### **Step 3: Forming the Lagrangian** To find the Lagrange dual, introduce Lagrange multipliers $$\lambda_1 \geq 0$$ and $$\lambda_2 \geq 0$$ for the two inequality constraints. The Lagrangian is: $$\mathcal{L}(x_1, x_2, x_3, \lambda_1, \lambda_2) = f(x) + \lambda_1(1 - x_1 - x_2 - x_3) + \lambda_2(1 - x_3)$$ Substitute $$f(x)$$: $$\mathcal{L}(x_1, x_2, x_3, \lambda_1, \lambda_2) = \frac{1}{2}x_1^2 + \frac{1}{2}x_2^2 + 2x_1x_2 + x_1 - x_2 - x_3 + \lambda_1(1 - x_1 - x_2 - x_3) + \lambda_2(1 - x_3)$$ $$\mathcal{L}(x_1, x_2, x_3, \lambda_1, \lambda_2) = \frac{1}{2}x_1^2 + \frac{1}{2}x_2^2 + 2x_1x_2 + x_1 - x_2 - x_3 - \lambda_1x_1 - \lambda_1x_2 - \lambda_1x_3 - \lambda_2x_3 + \lambda_1 + \lambda_2$$ --- ### **Step 4: Dual Function** The dual function $$g(\lambda_1, \lambda_2)$$ is obtained by minimizing $$\mathcal{L}$$ over $$x_1, x_2, x_3$$: $$g(\lambda_1, \lambda_2) = \inf_{x_1, x_2, x_3} \mathcal{L}(x_1, x_2, x_3, \lambda_1, \lambda_2)$$ #### Differentiate w.r.t $$x_1, x_2, x_3$$: 1. $$\frac{\partial \mathcal{L}}{\partial x_1} = x_1 + 2x_2 + 1 - \lambda_1 = 0 \implies x_1 = \lambda_1 - 2x_2 - 1$$ 2. $$\frac{\partial \mathcal{L}}{\partial x_2} = x_2 + 2x_1 - 1 - \lambda_1 = 0 \implies x_2 = (1 + \lambda_1 - 2x_1)/2$$ 3. $$\frac{\partial \mathcal{L}}{\partial x_3} = -1 - \lambda_1 - \lambda_2 = 0 \implies \lambda_1 + \lambda_2 = -1$$ --- This results in a set of equations to solve. We substitute back to find $$g(\lambda_1, \lambda_2)$$. After simplifying, the dual is derived as the maximum of $$g$$, subject to the dual feasibility conditions ($$\lambda_1, \lambda_2 \geq 0$$). If you'd like, I can proceed further with numerical simplifications or graphical representation of the dual.