Problem 175 - Advanced Calculus

Question

Find a relative minimum for \(\mathrm{f}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\) $$ 2 x^{2}+2 y^{2}+2 z^{2}-2 x z-2 y z-6 x+2 y+8 z+14 $$

Step-by-Step Solution

Answer

The function \(f(x, y, z) = 2x^{2} + 2y^{2} + 2z^{2} - 2xz - 2yz - 6x + 2y + 8z + 14\) has a relative minimum at the point (1, 1, 2).

Step 1: Find the gradient of the function

Find the gradient of the function f(x, y, z) by taking the partial derivative with respect to each variable (x, y, z). \[ \nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right) \] \[ \frac{\partial f}{\partial x} = 4x - 2z - 6 \\ \frac{\partial f}{\partial y} = 4y - 2z + 2 \\ \frac{\partial f}{\partial z} = 4z - 2x - 2y + 8 \] So, the gradient is: \[ \nabla f = \left(4x - 2z - 6, 4y - 2z + 2, 4z - 2x - 2y + 8\right) \]

Step 2: Find the critical points

Set the gradient to the zero vector and solve for x, y, and z: \[ 4x - 2z - 6 = 0 \\ 4y - 2z + 2 = 0 \\ 4z - 2x - 2y + 8 = 0 \] Solving this system of equations, we get: \(x = 1\), \(y = 1\), and \(z = 2\) Thus, the critical point is at (1, 1, 2).

Step 3: Apply the Second Partial Derivative Test

Calculate the second partial derivatives of f: \[ \frac{\partial^2 f}{\partial x^2} = 4 \\ \frac{\partial^2 f}{\partial y^2} = 4 \\ \frac{\partial^2 f}{\partial z^2} = 4 \\ \frac{\partial^2 f}{\partial x\partial y} = 0 \\ \frac{\partial^2 f}{\partial x\partial z} = -2 \\ \frac{\partial^2 f}{\partial y\partial z} = -2 \\ \] The Hessian matrix H is given by: \[ H = \begin{pmatrix} 4 & 0 & -2 \\ 0 & 4 & -2 \\ -2 & -2 & 4 \end{pmatrix} \] Evaluate the determinant of the matrix: \[ \text{det}(H) = 4(4)(4 - 2(-2)) - (-2)(-2)(-2) = 64 - 8 = 56 \] Since the determinant is positive and the second partial derivative with respect to x (\(\frac{\partial^2 f}{\partial x^2}\)) is positive, the Second Partial Derivative Test indicates a relative minimum. So, the function f(x, y, z) has a relative minimum at the point (1, 1, 2).

Key Concepts

Gradient of a Function

The gradient of a function is a vector that consists of the partial derivatives with respect to each variable. It provides the direction of the steepest ascent of the function. For a function like \( f(x, y, z) \), the gradient is denoted by \( abla f \) and is calculated as

• \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \).

This calculation provides a tool to analyze multidimensional functions. It helps in determining where the function is increasing or decreasing most rapidly. In the context of optimization, critical points often occur where the gradient is zero or undefined.

Critical Points

Critical points are the values of \( x, y, z \) where the gradient of the function equals zero. They provide potential locations for maxima, minima, or saddle points of the function. Finding critical points involves setting each component of the gradient to zero and solving the resulting equations. Here, the equations:

• \( 4x - 2z - 6 = 0 \)
• \( 4y - 2z + 2 = 0 \)
• \( 4z - 2x - 2y + 8 = 0 \)

When you solve these, you find \( x = 1, y = 1, z = 2 \). Thus, the critical point for the function \( f(x, y, z) \) is at \((1, 1, 2)\).

Partial Derivatives

Partial derivatives are derivatives of a multivariable function with respect to one variable, treating the others as constants. They are essential for understanding changes in a function along each coordinate axis. For example, in the function \( f(x, y, z) = 2x^2 + 2y^2 + 2z^2 - 2xz - 2yz - 6x + 2y + 8z + 14 \), the partial derivatives are:

• \( \frac{\partial f}{\partial x} = 4x - 2z - 6 \)
• \( \frac{\partial f}{\partial y} = 4y - 2z + 2 \)
• \( \frac{\partial f}{\partial z} = 4z - 2x - 2y + 8 \)

Each partial derivative represents how the function changes as one specific variable changes, while the others remain constant.

Second Partial Derivative Test

The Second Partial Derivative Test is employed to determine the nature of critical points in a function of multiple variables. It involves computing second order partial derivatives and evaluating the Hessian determinant. Here, the second partial derivatives are

• \( \frac{\partial^2 f}{\partial x^2} = 4 \)
• \( \frac{\partial^2 f}{\partial y^2} = 4 \)
• \( \frac{\partial^2 f}{\partial z^2} = 4 \)
• Mixed derivatives: \( \frac{\partial^2 f}{\partial x \partial z} = -2 \)

For the Hessian matrix \( H \), when its determinant is positive and one of the principal minors (like \( \frac{\partial^2 f}{\partial x^2} \)) is positive, the critical point is a relative minimum.

Hessian Matrix

The Hessian Matrix is a square matrix of second-order partial derivatives of a function. It provides information about the local curvature of a function. For our function, the Hessian is\[H = \begin{pmatrix} 4 & 0 & -2 \ 0 & 4 & -2 \ -2 & -2 & 4 \end{pmatrix}\]Calculating its determinant helps us apply the Second Partial Derivative Test. The determinant here is calculated as 56, which is positive.A positive determinant, in conjunction with positive second partial derivatives, confirms the presence of a relative minimum at the critical point. Hence, understanding the Hessian is crucial in optimizing multivariable functions.