Let's solve the problem step by step: ## Given Function \[ f(x, y, z) = x^2 + 2y^2 + 2z^2 + 2z^2 + 2xy + 2yz \] Let's simplify: \[ f(x, y, z) = x^2 + 2y^2 + 4z^2 + 2xy + 2yz \] --- ### **a) Calculate the gradient vector and find the stationary point(s).** The gradient vector is: \[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \] Calculate partial derivatives: #### \( \frac{\partial f}{\partial x} \): \[ \frac{\partial f}{\partial x} = 2x + 2y \] #### \( \frac{\partial f}{\partial y} \): \[ \frac{\partial f}{\partial y} = 4y + 2x + 2z \] #### \( \frac{\partial f}{\partial z} \): \[ \frac{\partial f}{\partial z} = 8z + 2y \] So, \[ \nabla f = (2x + 2y,\, 4y + 2x + 2z,\, 8z + 2y) \] **Stationary points** occur when all components are zero: \[ \begin{cases} 2x + 2y = 0 \\ 4y + 2x + 2z = 0 \\ 8z + 2y = 0 \\ \end{cases} \] Let's solve this system: 1. \(2x + 2y = 0 \implies x = -y\) 2. \(8z + 2y = 0 \implies 4z + y = 0 \implies y = -4z\) 3. Substitute \(x = -y\) and \(y = -4z\) into \(4y + 2x + 2z = 0\): \[ 4y + 2x + 2z = 0 \\ 4(-4z) + 2(-(-4z)) + 2z = 0 \\ -16z + 8z + 2z = 0 \\ (-16z + 8z + 2z) = (-16+8+2)z = -6z = 0 \implies z = 0 \] So, \(z = 0\), \(y = -4z = 0\), \(x = -y = 0\). **Stationary point:** \(\boxed{(0, 0, 0)}\) --- ### **b) Compute the Hessian matrix at the stationary point and determine its sign definiteness.** The Hessian matrix \(\mathbf{H}\) is the matrix of second derivatives: #### Compute second derivatives: - \(f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2\) - \(f_{yy} = \frac{\partial^2 f}{\partial y^2} = 4\) - \(f_{zz} = \frac{\partial^2 f}{\partial z^2} = 8\) - \(f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 2\) - \(f_{xz} = \frac{\partial^2 f}{\partial x \partial z} = 0\) - \(f_{yz} = \frac{\partial^2 f}{\partial y \partial z} = 2\) The Hessian is: \[ H = \begin{bmatrix} 2 & 2 & 0 \\ 2 & 4 & 2 \\ 0 & 2 & 8 \\ \end{bmatrix} \] #### **Sign definiteness (Sylvester's criterion):** Calculate leading principal minors: 1. First minor: \(2 > 0\) 2. Second order minor: \[ \begin{vmatrix} 2 & 2 \\ 2 & 4 \end{vmatrix} = (2)(4) - (2)(2) = 8 - 4 = 4 > 0 \] 3. Third order determinant: \[ \begin{vmatrix} 2 & 2 & 0 \\ 2 & 4 & 2 \\ 0 & 2 & 8 \\ \end{vmatrix} = 2 \begin{vmatrix} 4 & 2 \\ 2 & 8 \end{vmatrix} - 2 \begin{vmatrix} 2 & 2 \\ 0 & 8 \end{vmatrix} + 0 \cdot D \] Calculate: - \(\begin{vmatrix} 4 & 2 \\ 2 & 8 \end{vmatrix} = 4 \cdot 8 - 2 \cdot 2 = 32 - 4 = 28\) - \(\begin{vmatrix} 2 & 2 \\ 0 & 8 \end{vmatrix} = 2 \cdot 8 - 2 \cdot 0 = 16\) So, \[ = 2 \cdot 28 - 2 \cdot 16 = 56 - 32 = 24 > 0 \] All leading principal minors are positive, so the Hessian is **positive definite** at \((0, 0, 0)\). --- ### **Conclusion** - **Stationary point:** \(\boxed{(0, 0, 0)}\) - **Type:** Since the Hessian is positive definite, this is a **local minimum**. --- #### **Final Answers** **a)** Gradient vector: \(\boxed{(2x + 2y,\, 4y + 2x + 2z,\, 8z + 2y)}\), Stationary point: \(\boxed{(0, 0, 0)}\). **b)** Hessian: \[ \boxed{ \begin{bmatrix} 2 & 2 & 0 \\ 2 & 4 & 2 \\ 0 & 2 & 8 \\ \end{bmatrix} } \] At \((0, 0, 0)\): **positive definite**, so the stationary point is a \(\boxed{\text{local minimum}}\).