91Ó°ÊÓ

First, we compute the gradients of the function \(f\) and the constraint function \(g\): For the function \(f(x,y,z) = x^2 + xz + y^2\), we have $$ \nabla f(x,y,z) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right) = (2x+z, 2y, x). $$ For the constraint \(g(x,y,z) = x + y + z - 1\), we have $$ \nabla g(x,y,z) = \left(\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}, \frac{\partial g}{\partial z}\right) = (1, 1, 1). $$

Now, we set up the system of equations using the method of Lagrange multipliers: $$ \nabla f(x,y,z) = \lambda \nabla g(x,y,z) \implies (2x+z, 2y, x) = (\lambda, \lambda, \lambda) $$ and the constraint equation \(g(x,y,z)=0 \implies x + y + z - 1 = 0\).

From the system of equations in Step 2, we can write $$ \begin{cases} 2x+z=\lambda \\ 2y=\lambda \\ x=\lambda \\ x+y+z=1 \end{cases} $$ Using the third equation, we have \(x = \lambda\). Substituting this into the first equation gives $$ 2\lambda + z = \lambda \implies z = -\lambda. $$ Substituting the expressions for x and z into the constraint equation: $$ \lambda - \lambda + (-\lambda) = 1 \implies -\lambda = 1 \implies \lambda = -1. $$ Then, \(x=\lambda = -1\), and \(z=-\lambda = 1\). Using the second equation, we have \(2y = \lambda = -1\), and thus, \(y = -\frac{1}{2}\).

From the solution in Step 3, we found that \((x, y, z) = (-1, -\frac{1}{2}, 1)\) is a candidate stationary point. Now we need to check if it corresponds to a maximum or a minimum of function \(f\).

To analyze the critical point, we compute the Hessian matrix of the function \(f\): $$ H_f(x,y,z) = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x\partial y} & \frac{\partial^2 f}{\partial x\partial z} \\ \frac{\partial^2 f}{\partial y\partial x} & \frac{\partial^2 f}{\partial y^2} & \frac{\partial^2 f}{\partial y\partial z} \\ \frac{\partial^2 f}{\partial z\partial x} & \frac{\partial^2 f}{\partial z\partial y} & \frac{\partial^2 f}{\partial z^2} \end{bmatrix}= \begin{bmatrix} 2 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & 0 & 0 \end{bmatrix} $$ Now we compute the eigenvalues of the Hessian matrix: $$ \det(H_f - \lambda I) = \begin{vmatrix} 2-\lambda & 0 & 1 \\ 0 & 2-\lambda & 0 \\ 1 & 0 & -\lambda \end{vmatrix} = \lambda^{3} - 4\lambda^{2} + 4\lambda = \lambda(\lambda^2 -4\lambda +4)=\lambda(\lambda-2)^2 $$ The eigenvalues are \(\lambda_1 = 0\), \(\lambda_2 = 2\), and \(\lambda_3 = 2\). Since there's an eigenvalue equal to zero, we cannot determine if the stationary point is a maximum or a minimum using the Hessian matrix. In conclusion, there is a stationary point at \((-1, -\frac{1}{2}, 1)\), but we cannot determine if it is an extremum using the Hessian matrix method.