Solve the following problem (check the NDCQ)
max 2 +  + 42 subject to ()
2+2≤1 ≥0 and≥0

To solve this optimization problem, we'll proceed with the following steps:

1. **Objective Function:**
   The objective function to maximize is:
   \[
   f(\xi, \psi) = \xi^2 + \xi + 4\psi^2
   \]

2. **Constraints:**
   The constraints are given by:
   \[
   2\xi + 2\psi \leq 1 \quad \text{(1)}
   \]
   \[
   \xi \geq 0 \quad \text{(2)}
   \]
   \[
   \psi \geq 0 \quad \text{(3)}
   \]

3. **Non-negativity Condition Qualification (NDCQ):**
   To check whether the problem satisfies the Non-negativity Condition Qualification (NDCQ), we need to verify that the gradients of the active constraints at the optimal solution are not linearly dependent.

### Step-by-Step Solution:

**1. Setting up the Lagrangian:**

   First, we define the Lagrangian for this problem:
   \[
   \mathcal{L}(\xi, \psi, \lambda) = \xi^2 + \xi + 4\psi^2 + \lambda (1 - 2\xi - 2\psi)
   \]
   Here, \(\lambda\) is the Lagrange multiplier associated with the constraint \(2\xi + 2\psi \leq 1\).

**2. KKT Conditions:**

   The Karush-Kuhn-Tucker (KKT) conditions are necessary for optimality. The conditions are:

   - Stationarity:
     \[
     \frac{\partial \mathcal{L}}{\partial \xi} = 2\xi + 1 - 2\lambda = 0 \quad \Rightarrow \quad \xi = \frac{2\lambda - 1}{2}
     \]
     \[
     \frac{\partial \mathcal{L}}{\partial \psi} = 8\psi - 2\lambda = 0 \quad \Rightarrow \quad \psi = \frac{\lambda}{4}
     \]

   - Primal feasibility:
     \[
     2\xi + 2\psi \leq 1, \quad \xi \geq 0, \quad \psi \geq 0
     \]

   - Dual feasibility:
     \[
     \lambda \geq 0
     \]

   - Complementary slackness:
     \[
     \lambda (1 - 2\xi - 2\psi) = 0
     \]

**3. Solving the KKT Conditions:**

   From the stationarity condition, we have:
   \[
   \xi = \frac{2\lambda - 1}{2}, \quad \psi = \frac{\lambda}{4}
   \]

   Substitute these into the constraint \(2\xi + 2\psi \leq 1\):
   \[
   2\left(\frac{2\lambda - 1}{2}\right) + 2\left(\frac{\lambda}{4}\right) \leq 1
   \]
   Simplifying:
   \[
   2\lambda - 1 + \frac{\lambda}{2} \leq 1
   \]
   \[
   \frac{5\lambda}{2} \leq 2 \quad \Rightarrow \quad \lambda \leq \frac{4}{5}
   \]

   The value of \(\lambda\) should also be non-negative, so \(0 \leq \lambda \leq \frac{4}{5}\).

**4. Checking the values of \(\xi\) and \(\psi\) at \(\lambda = \frac{4}{5}\):**

   - For \(\xi\):
     \[
     \xi = \frac{2\left(\frac{4}{5}\right) - 1}{2} = \frac{\frac{8}{5} - 1}{2} = \frac{3}{10}
     \]

   - For \(\psi\):
     \[
     \psi = \frac{\frac{4}{5}}{4} = \frac{1}{5}
     \]

   Check the constraint \(2\xi + 2\psi \leq 1\):
   \[
   2 \cdot \frac{3}{10} + 2 \cdot \frac{1}{5} = \frac{6}{10} + \frac{2}{5} = 1
   \]
   which satisfies the inequality.

**5. Objective Function Value at \(\xi = \frac{3}{10}\), \(\psi = \frac{1}{5}\):**

   \[
   f\left(\frac{3}{10}, \frac{1}{5}\right) = \left(\frac{3}{10}\right)^2 + \frac{3}{10} + 4\left(\frac{1}{5}\right)^2
   \]
   \[
   = \frac{9}{100} + \frac{30}{100} + \frac{16}{25}
   \]
   \[
   = \frac{39}{100} + \frac{64}{100}
   \]
   \[
   = \frac{103}{100} = 1.03
   \]

Therefore, the maximum value of the objective function \( \xi^2 + \xi + 4\psi^2 \) is **1.03**, attained at \(\xi = \frac{3}{10}\) and \(\psi = \frac{1}{5}\).

### Summary:
- Maximum value of the objective function: **1.03**
- Optimal values: \(\xi = \frac{3}{10}\), \(\psi = \frac{1}{5}\)

### Questions and Tip:
1. What are the KKT conditions, and why are they important?
2. How do you interpret Lagrange multipliers in optimization problems?
3. Can we have multiple optimal solutions in non-linear optimization problems?
4. How does the presence of constraints affect the solution to optimization problems?
5. What would happen if we changed the objective function or constraints?

**Tip:** Always check if the constraints are active or inactive at the solution point to understand the behavior of the optimization problem better.

Learn how to solve the optimization problem to maximize the objective function \(\xi^2 + \xi + 4\psi^2\) subject to constraints. Detailed step-by-step solution using Lagrangian multipliers and KKT conditions. Suitable for advanced undergraduate students and beyond.

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