Chapter 3: Problem 1
The function $$ f\left(x_{1}, x_{2}, x_{3}\right)=x_{1}^{2}+x_{2}^{2}+3 x_{3}^{2}-x_{1} x_{2}+2 x_{1} x_{3}+x_{2} x_{3} $$ defined on \(R^{3}\) has only one stationary point. Show that it is a local minimum point.
Short Answer
Expert verified
The stationary point \( (0, 0, 0) \) is a local minimum because the Hessian matrix is positive definite.
Step by step solution
01
- Compute the Partial Derivatives of f
To find the stationary points, calculate the partial derivatives with respect to each variable:\( f_{x_1} = \frac{\partial f}{\partial x_1} = 2x_1 - x_2 + 2x_3 \)\( f_{x_2} = \frac{\partial f}{\partial x_2} = 2x_2 - x_1 + x_3 \)\( f_{x_3} = \frac{\partial f}{\partial x_3} = 6x_3 + 2x_1 + x_2 \)
02
- Determine the Stationary Point
Set the partial derivatives to zero and solve the system of equations:\( 2x_1 - x_2 + 2x_3 = 0 \)\( 2x_2 - x_1 + x_3 = 0 \)\( 6x_3 + 2x_1 + x_2 = 0 \)Solving this system yields the stationary point \( (0, 0, 0) \).
03
- Compute the Hessian Matrix
The Hessian matrix \( H \) is given by the second partial derivatives:\( H = \begin{bmatrix} \frac{\partial^2 f}{\partial x_1^2} & \frac{\partial^2 f}{\partial x_1 \partial x_2} & \frac{\partial^2 f}{\partial x_1 \partial x_3} \ \frac{\partial^2 f}{\partial x_2 \partial x_1} & \frac{\partial^2 f}{\partial x_2^2} & \frac{\partial^2 f}{\partial x_2 \partial x_3} \ \frac{\partial^2 f}{\partial x_3 \partial x_1} & \frac{\partial^2 f}{\partial x_3 \partial x_2} & \frac{\partial^2 f}{\partial x_3^2} \end{bmatrix} \)Calculate each element:\( H = \begin{bmatrix} 2 & -1 & 2 \ -1 & 2 & 1 \ 2 & 1 & 6 \end{bmatrix} \)
04
- Determine Positive Definiteness of the Hessian
A local minimum occurs if the Hessian is positive definite. Compute the leading principal minors:\( D_1 = 2 \)\( D_2 = \begin{vmatrix} 2 & -1 \ -1 & 2 \end{vmatrix} = 3 \)\( D_3 = \begin{vmatrix} 2 & -1 & 2 \ -1 & 2 & 1 \ 2 & 1 & 6 \end{vmatrix} = 35 \)All leading principal minors \( D_1, D_2, \) and \( D_3 \) are positive, indicating the Hessian is positive definite.
05
- Conclusion
Since the Hessian matrix is positive definite at the stationary point \( (0, 0, 0) \), this point is a local minimum.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
partial derivatives
Partial derivatives are a fundamental tool in multivariable calculus that help us understand how a function changes with respect to one of its variables, while keeping the other variables constant. For a function of three variables, such as \(f(x_1, x_2, x_3)\), we compute the partial derivatives with respect to \(x_1, x_2,\) and \(x_3\) to explore the function's behavior. Specifically, we find the partial derivatives:
- \(f_{x_1} = \frac{\partial f}{\partial x_1}\)
- \(f_{x_2} = \frac{\partial f}{\partial x_2}\)
- \(f_{x_3} = \frac{\partial f}{\partial x_3}\)
- \(f_{x_1} = 2x_1 - x_2 + 2x_3\)
- \(f_{x_2} = 2x_2 - x_1 + x_3\)
- \(f_{x_3} = 6x_3 + 2x_1 + x_2\)
stationary point
A stationary point is where the partial derivatives of a function are zero. This means the function does not change in any direction from this point. To find a stationary point, we solve the system of equations obtained by setting the partial derivatives to zero:
- \(2x_1 - x_2 + 2x_3 = 0\)
- \(2x_2 - x_1 + x_3 = 0\)
- \(6x_3 + 2x_1 + x_2 = 0\)
Hessian matrix
The Hessian matrix is a square matrix of second-order partial derivatives of a function. It provides essential information about the curvature of the function at a stationary point. For a function of three variables, the Hessian matrix \(H\) is:
\[ H = \begin{bmatrix} \frac{\partial^2 f}{\partial x_1^2} & \frac{\partial^2 f}{\partial x_1 \partial x_2} & \frac{\partial^2 f}{\partial x_1 \partial x_3} \ \frac{\partial^2 f}{\partial x_2 \partial x_1} & \frac{\partial^2 f}{\partial x_2^2} & \frac{\partial^2 f}{\partial x_2 \partial x_3} \ \frac{\partial^2 f}{\partial x_3 \partial x_1} & \frac{\partial^2 f}{\partial x_3 \partial x_2} & \frac{\partial^2 f}{\partial x_3^2} \end{bmatrix} \]
For our function, the Hessian matrix is:
\ H = \begin{bmatrix} 2 & -1 & 2 \ -1 & 2 & 1 \ 2 & 1 & 6 \end{bmatrix} \
We use the Hessian matrix to determine if a stationary point is a local minimum, maximum, or saddle point.
\[ H = \begin{bmatrix} \frac{\partial^2 f}{\partial x_1^2} & \frac{\partial^2 f}{\partial x_1 \partial x_2} & \frac{\partial^2 f}{\partial x_1 \partial x_3} \ \frac{\partial^2 f}{\partial x_2 \partial x_1} & \frac{\partial^2 f}{\partial x_2^2} & \frac{\partial^2 f}{\partial x_2 \partial x_3} \ \frac{\partial^2 f}{\partial x_3 \partial x_1} & \frac{\partial^2 f}{\partial x_3 \partial x_2} & \frac{\partial^2 f}{\partial x_3^2} \end{bmatrix} \]
For our function, the Hessian matrix is:
\ H = \begin{bmatrix} 2 & -1 & 2 \ -1 & 2 & 1 \ 2 & 1 & 6 \end{bmatrix} \
We use the Hessian matrix to determine if a stationary point is a local minimum, maximum, or saddle point.
positive definite
A matrix is positive definite if all its leading principal minors are positive. This property is crucial in establishing whether a stationary point is a local minimum. For our Hessian matrix, we calculate the leading principal minors:
- First minor: \(D_1 = 2\)
- Second minor: \(D_2 = \begin{vmatrix} 2 & -1 \ -1 & 2 \end{vmatrix} = 3\)
- Third minor: \(D_3 = \begin{vmatrix} 2 & -1 & 2 \ -1 & 2 & 1 \ 2 & 1 & 6 \end{vmatrix} = 35\)