Let's solve **Problem (i):** ### \( f(x_1, x_2) = (4x_1^2 - x_2)^2 \) #### **Step 1: Find the stationary points** Stationary points occur where the gradient is zero: \[ \nabla f = \begin{bmatrix} \frac{\partial f}{\partial x_1} \\ \frac{\partial f}{\partial x_2} \end{bmatrix} \] First, compute the partial derivatives: - \( \frac{\partial f}{\partial x_1} = 2(4x_1^2 - x_2) \cdot (8x_1) = 16x_1(4x_1^2 - x_2) \) - \( \frac{\partial f}{\partial x_2} = 2(4x_1^2 - x_2) \cdot (-1) = -2(4x_1^2 - x_2) \) Set the gradient to zero: \[ \begin{cases} 16x_1(4x_1^2 - x_2) = 0 \\ -2(4x_1^2 - x_2) = 0 \end{cases} \] From the second equation: \[ -2(4x_1^2 - x_2) = 0 \implies 4x_1^2 - x_2 = 0 \implies x_2 = 4x_1^2 \] Substitute into the first equation: \[ 16x_1(4x_1^2 - x_2) = 16x_1(4x_1^2 - 4x_1^2) = 16x_1(0) = 0 \] So, **for any \( x_1 \), \( x_2 = 4x_1^2 \) is a stationary point**. #### **Step 2: Classify the stationary points** Let's compute the Hessian matrix: - \( f(x_1, x_2) = (4x_1^2 - x_2)^2 \) Second derivatives: - \( f_{x_1x_1} = \frac{\partial}{\partial x_1} \left( 16x_1(4x_1^2 - x_2) \right) = 16(4x_1^2 - x_2) + 16x_1 \cdot 8x_1 = 16(4x_1^2 - x_2) + 128x_1^2 = 16(4x_1^2 - x_2 + 8x_1^2) = 16(12x_1^2 - x_2) \) - \( f_{x_1x_2} = \frac{\partial}{\partial x_2} \left( 16x_1(4x_1^2 - x_2) \right) = -16x_1 \) - \( f_{x_2x_1} = \frac{\partial}{\partial x_1} \left( -2(4x_1^2 - x_2) \right) = -2 \cdot 8x_1 = -16x_1 \) - \( f_{x_2x_2} = \frac{\partial}{\partial x_2} \left( -2(4x_1^2 - x_2) \right) = 2 \) So the Hessian is: \[ H = \begin{bmatrix} 16(12x_1^2 - x_2) & -16x_1 \\ -16x_1 & 2 \end{bmatrix} \] At stationary points, \( x_2 = 4x_1^2 \), so substitute this in: - \( 12x_1^2 - x_2 = 12x_1^2 - 4x_1^2 = 8x_1^2 \) - So \( f_{x_1x_1} = 16 \cdot 8x_1^2 = 128x_1^2 \) - \( f_{x_1x_2} = f_{x_2x_1} = -16x_1 \) - \( f_{x_2x_2} = 2 \) So, \[ H = \begin{bmatrix} 128x_1^2 & -16x_1 \\ -16x_1 & 2 \end{bmatrix} \] Let's check definiteness. The principal minors are: - First minor: \( 128x_1^2 \geq 0 \) - Determinant: \( (128x_1^2)(2) - (-16x_1)^2 = 256x_1^2 - 256x_1^2 = 0 \) So, **the Hessian is positive semi-definite** (not positive definite unless \( x_1 \neq 0 \)), but determinant is zero everywhere. **Check the function value:** \[ f(x_1, x_2) = (4x_1^2 - x_2)^2 = (0)^2 = 0 \] This is the global minimum, since the function is always non-negative. ### **Conclusion for (i):** - **Stationary points:** All points of the form \((x_1, x_2) = (a, 4a^2)\), \(a \in \mathbb{R}\) - **Type:** All these points are **global minima** (but not strict unless \(x_1 = 0\)). The set is a parabola in \( (x_1, x_2) \) space. --- Would you like to proceed with part (ii), or do you have questions on the above?

Let's analyze **Problem (ii):** ### \( f(x_1, x_2) = 2x_1^3 - 6x_1 x_2 + 3x_2^2 \) --- #### **Step 1: Find the stationary points** Compute the gradient: \[ \nabla f = \begin{bmatrix} \frac{\partial f}{\partial x_1} \\ \frac{\partial f}{\partial x_2} \end{bmatrix} \] Calculate each partial derivative: - \( \frac{\partial f}{\partial x_1} = 6x_1^2 - 6x_2 \) - \( \frac{\partial f}{\partial x_2} = -6x_1 + 6x_2 \) Set the gradient to zero: \[ \begin{cases} 6x_1^2 - 6x_2 = 0 \\ -6x_1 + 6x_2 = 0 \end{cases} \] From the second equation: \[ -6x_1 + 6x_2 = 0 \implies x_2 = x_1 \] Substitute into the first: \[ 6x_1^2 - 6x_1 = 0 \implies 6x_1 (x_1 - 1) = 0 \] Solutions: \[ x_1 = 0 \quad \text{or} \quad x_1 = 1 \] Corresponding \(x_2\): - If \(x_1=0\), then \(x_2=0\) - If \(x_1=1\), then \(x_2=1\) **Stationary points:** \[ \boxed{ (0,0) \quad \text{and} \quad (1,1) } \] --- #### **Step 2: Classify the stationary points** Compute the Hessian matrix: \[ H = \begin{bmatrix} \frac{\partial^2 f}{\partial x_1^2} & \frac{\partial^2 f}{\partial x_1 \partial x_2} \\ \frac{\partial^2 f}{\partial x_2 \partial x_1} & \frac{\partial^2 f}{\partial x_2^2} \end{bmatrix} \] Calculate second derivatives: - \( f_{x_1x_1} = 12x_1 \) - \( f_{x_1x_2} = -6 \) - \( f_{x_2x_2} = 6 \) So, \[ H = \begin{bmatrix} 12x_1 & -6 \\ -6 & 6 \end{bmatrix} \] Evaluate at each stationary point: --- **At \((0,0)\):** \[ H = \begin{bmatrix} 0 & -6 \\ -6 & 6 \end{bmatrix} \] Calculate the leading principal minors: - \( D_1 = 0 \) (zero, so the Hessian is singular) - \( D_2 = (0)(6) - (-6)(-6) = 0 - 36 = -36 < 0 \) Since the determinant is negative, the Hessian is indefinite at \((0,0)\). **Conclusion:** \((0,0)\) is a **saddle point**. --- **At \((1,1)\):** \[ H = \begin{bmatrix} 12 \times 1 & -6 \\ -6 & 6 \end{bmatrix} = \begin{bmatrix} 12 & -6 \\ -6 & 6 \end{bmatrix} \] Calculate minors: - \( D_1 = 12 > 0 \) - \( D_2 = (12)(6) - (-6)(-6) = 72 - 36 = 36 > 0 \) Since both minors are positive, the Hessian is **positive definite** at \((1,1)\). **Conclusion:** \((1,1)\) is a **strict local minimum**. --- ### **Final Classification:** | Stationary Point | Nature | |------------------|--------------------------| | \((0,0)\) | Saddle point | | \((1,1)\) | Strict local minimum | --- **Summary:** - **Stationary points:** \((0,0)\) (saddle), \((1,1)\) (minimum) - The point \((1,1)\) is a local minimum because the Hessian is positive definite there. - The point \((0,0)\) is a saddle point because the Hessian is indefinite. --- Would you like me to proceed with the next functions?