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

To solve the problem and check the Necessary and Sufficient Conditions for Quasi-concavity (NDCQ), let's follow these steps:

### Problem Statement
We are asked to maximize the objective function:
\[
f(\xi, \psi) = \xi^2 + \xi + 4\psi^2
\]
subject to the constraints:
\[
2\xi + 2\psi \leq 1, \quad \xi \geq 0, \quad \psi \geq 0.
\]

### 1. **Formulate the Lagrangian:**

The Lagrangian for this problem is given by:
\[
\mathcal{L}(\xi, \psi, \lambda) = \xi^2 + \xi + 4\psi^2 + \lambda(1 - 2\xi - 2\psi)
\]
where \(\lambda \geq 0\) is the Lagrange multiplier associated with the constraint \(2\xi + 2\psi \leq 1\).

### 2. **Compute the First-Order Conditions (FOCs):**

Take the partial derivatives of \(\mathcal{L}\) with respect to \(\xi\), \(\psi\), and \(\lambda\), and set them equal to zero.

\[
\frac{\partial \mathcal{L}}{\partial \xi} = 2\xi + 1 - 2\lambda = 0 \quad \Rightarrow \quad \lambda = \frac{2\xi + 1}{2}
\]

\[
\frac{\partial \mathcal{L}}{\partial \psi} = 8\psi - 2\lambda = 0 \quad \Rightarrow \quad \lambda = 4\psi
\]

\[
\frac{\partial \mathcal{L}}{\partial \lambda} = 1 - 2\xi - 2\psi \geq 0, \quad \lambda(1 - 2\xi - 2\psi) = 0
\]

### 3. **Solve the System of Equations:**

From \(\lambda = \frac{2\xi + 1}{2}\) and \(\lambda = 4\psi\), equate them to solve for \(\xi\) and \(\psi\):

\[
\frac{2\xi + 1}{2} = 4\psi \quad \Rightarrow \quad 2\xi + 1 = 8\psi \quad \Rightarrow \quad \xi = 4\psi - \frac{1}{2}
\]

Substitute \(\xi = 4\psi - \frac{1}{2}\) into the constraint \(2\xi + 2\psi \leq 1\):

\[
2(4\psi - \frac{1}{2}) + 2\psi \leq 1
\]
\[
8\psi - 1 + 2\psi \leq 1 \quad \Rightarrow \quad 10\psi \leq 2 \quad \Rightarrow \quad \psi \leq \frac{1}{5}
\]

Since \(\xi = 4\psi - \frac{1}{2}\) and \(\xi \geq 0\), then:

\[
4\psi - \frac{1}{2} \geq 0 \quad \Rightarrow \quad \psi \geq \frac{1}{8}
\]

Thus, \(\frac{1}{8} \leq \psi \leq \frac{1}{5}\).

### 4. **Evaluate the Objective Function:**

We will now evaluate the objective function \(f(\xi, \psi)\) for the boundary points:

- **Case 1**: \(\psi = \frac{1}{8}\)
  \[
  \xi = 4\left(\frac{1}{8}\right) - \frac{1}{2} = \frac{1}{2} - \frac{1}{2} = 0
  \]
  \[
  f(0, \frac{1}{8}) = 0^2 + 0 + 4\left(\frac{1}{8}\right)^2 = \frac{1}{16}
  \]

- **Case 2**: \(\psi = \frac{1}{5}\)
  \[
  \xi = 4\left(\frac{1}{5}\right) - \frac{1}{2} = \frac{4}{5} - \frac{1}{2} = \frac{3}{10}
  \]
  \[
  f\left(\frac{3}{10}, \frac{1}{5}\right) = \left(\frac{3}{10}\right)^2 + \frac{3}{10} + 4\left(\frac{1}{5}\right)^2 = \frac{9}{100} + \frac{3}{10} + \frac{4}{25} = \frac{9}{100} + \frac{30}{100} + \frac{16}{100} = \frac{55}{100} = \frac{11}{20}
  \]

Thus, the maximum value of the objective function is achieved at \(\psi = \frac{1}{5}\) and \(\xi = \frac{3}{10}\).

### **Conclusion:**
The maximum value of the objective function \(f(\xi, \psi) = \xi^2 + \xi + 4\psi^2\) is \(\frac{11}{20}\), subject to the constraints provided.

### NDCQ Check:

1. **Convexity of the constraint set**: The constraints form a convex set.
2. **Quasi-concavity of the objective function**: The Hessian matrix is not necessary since we're maximizing a quadratic function. We check for quasi-concavity through the sign of first-order derivatives. The first derivatives indicate a global maximum within the feasible region.

Would you like more details or have any questions?

### 5 Related Questions:
1. What is the significance of Lagrange multipliers in constrained optimization?
2. How do you

Learn how to maximize the function xi^2 + xi + 4*psi^2 subject to given constraints using Lagrange multipliers and the necessary and sufficient conditions for quasi-concavity (NDCQ). Explore detailed step-by-step solutions and mathematical insights.

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