Question

Find the extreme values, if any, of the following four functions. Check whether they are maxima or minima by the determinantal test. $$z=x_{1}^{2}+3 x_{2}^{2}-3 x_{1} x_{2}+4 x_{2} x_{3}+6 x_{3}^{2}$$

Short answer

The function has a minimum at the critical point \( x_1 = 0, x_2 = 0, x_3 = 0 \).

Step-by-step solution

Identify the Function and Variables

The given function is a multivariable quadratic function defined by \( z = x_1^2 + 3x_2^2 - 3x_1x_2 + 4x_2x_3 + 6x_3^2 \). The variables involved are \( x_1, x_2, \) and \( x_3 \). Our task is to find the extreme values of this function by determining the critical points and classifying them using the Hessian matrix.

Calculate the First Partial Derivatives

To find the critical points, we need to compute the first partial derivatives of the function with respect to each variable and set them to zero. The partial derivatives are: - \( \frac{\partial z}{\partial x_1} = 2x_1 - 3x_2 \)- \( \frac{\partial z}{\partial x_2} = 6x_2 - 3x_1 + 4x_3 \)- \( \frac{\partial z}{\partial x_3} = 4x_2 + 12x_3 \).Set each partial derivative equal to zero to find the critical points.

Solve the System for Critical Points

Set the partial derivatives to zero and solve the resulting system of equations:1. \( 2x_1 - 3x_2 = 0 \)2. \( 6x_2 - 3x_1 + 4x_3 = 0 \)3. \( 4x_2 + 12x_3 = 0 \).From equation (1), \( x_1 = \frac{3}{2}x_2 \). Substitute this into equations (2) and (3) to find the critical points. Simplifying gives one solution: \( x_1 = 0, x_2 = 0, x_3 = 0 \).

Compute the Hessian Matrix

The Hessian matrix consists of the second partial derivatives of the function. Calculate the second partial derivatives:- \( \frac{\partial^2 z}{\partial x_1^2} = 2 \), \( \frac{\partial^2 z}{\partial x_1 \partial x_2} = -3 \), \( \frac{\partial^2 z}{\partial x_1 \partial x_3} = 0 \)- \( \frac{\partial^2 z}{\partial x_2^2} = 6 \), \( \frac{\partial^2 z}{\partial x_2 \partial x_3} = 4 \)- \( \frac{\partial^2 z}{\partial x_3^2} = 12 \).The Hessian matrix is:\[H = \begin{bmatrix}2 & -3 & 0 \-3 & 6 & 4 \0 & 4 & 12\end{bmatrix}\]

Apply the Determinantal Test

Compute the leading principal minors of the Hessian matrix to determine the nature of the critical point. The minors are:1. \( M_1 = 2 \)2. \( M_2 = \begin{vmatrix} 2 & -3 \ -3 & 6 \end{vmatrix} = (2)(6) - (-3)(-3) = 12 - 9 = 3 \)3. \( M_3 = \begin{vmatrix} 2 & -3 & 0 \ -3 & 6 & 4 \ 0 & 4 & 12 \end{vmatrix} = 2(6 \times 12 - 4 \times 4) + 3(-3 \times 12 + 4 \times 0) = 2(72 - 16) - 36 = 112 - 36 = 76 \).Since all minors \( M_1 > 0, M_2 > 0, M_3 > 0 \), the function has a minimum at the critical point.

Key concepts

Hessian Matrix

The Hessian matrix is a powerful tool in mathematical optimization that helps us understand the nature of critical points in a multivariable function. You can think of it as a matrix composed of all the second-order partial derivatives of a function. In simple terms, it tells us how the function curves around its critical points.
To construct a Hessian matrix for a function with variables like \(x_1, x_2, x_3\), you need to calculate the second derivatives for each possible pair of variables:

• Second derivative with respect to the same variable, like \(\frac{\partial^2 z}{\partial x_1^2}\).
• Mixed second derivatives, like \(\frac{\partial^2 z}{\partial x_1 \partial x_2}\).

By combining these second derivatives in a specific order, we get the Hessian matrix, which in the case of our function is a 3x3 matrix.

Determinantal Test

The Determinantal Test uses the Hessian matrix to decide whether a given critical point is a minimum, maximum, or a saddle point. We do this by evaluating the determinants of the leading principal minors of the Hessian.
Here's how it works:

• Principal Minor: It's a smaller square matrix derived from the top-left corner of the Hessian.
• Check their determinants sequentially: \(M_1\), \(M_2\), and \(M_3\).
• The signs of these determinants tell us the nature of the critical point:
• All positive: Local minimum.
• Alternating in sign starting with a negative: Local maximum.
• Any other pattern: Saddle point.

For the function in our exercise, all principal minors were positive, indicating a local minimum.

Critical Points

Critical points are where the first derivatives (slopes) of the function become zero, representing where the function may have extreme values (like peaks, valleys, or saddle points).
To find these points for our function, we followed these steps:

• First Partial Derivatives: Compute these derivatives with respect to each variable and find where they all equal zero.
• Solve the resulting system of equations, leading you to the critical point.

In our problem, we found that \(x_1 = 0, x_2 = 0, x_3 = 0\) forms a critical point, meaning around this point, the function's slope is flat in all directions. The nature of this point was further analyzed using the Hessian matrix and Determinantal Test to confirm it's a minimum.