Question

Find all critical points. Indicate whether each such point gives a local maximum or a local minimum, or whether it is a saddle point. Hint: Use Theorem \(\mathrm{C} .\) \(f(x, y)=x^{2}+4 y^{2}-4 x\)

Short answer

The critical point is (2, 0) and it is a local minimum.

Step-by-step solution

Find the Gradient

To find critical points, calculate the gradient of the function \( f(x, y) = x^2 + 4y^2 - 4x \). The gradient is found by computing \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). Here, \( \frac{\partial f}{\partial x} = 2x - 4 \) and \( \frac{\partial f}{\partial y} = 8y \).

Set the Gradient to Zero

Set each component of the gradient to zero to find the critical points. Solve the equations \( 2x - 4 = 0 \) and \( 8y = 0 \). Solving these gives \( x = 2 \) and \( y = 0 \). Therefore, the critical point is \((2, 0)\).

Calculate the Hessian Matrix

The Hessian matrix \( H \) is used to classify critical points. It is the matrix of second partial derivatives: \( H = \begin{pmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} \end{pmatrix} \). Here, \( \frac{\partial^2 f}{\partial x^2} = 2 \), \( \frac{\partial^2 f}{\partial x \partial y} = 0 \), and \( \frac{\partial^2 f}{\partial y^2} = 8 \). So, \( H = \begin{pmatrix} 2 & 0 \ 0 & 8 \end{pmatrix} \).

Determine the Nature of Critical Points

The determinant of the Hessian \( \det(H) = 2 \times 8 - 0 \times 0 = 16 \) is positive, indicating that the test is inconclusive if both second derivatives were negative or positive. Since \( \frac{\partial^2 f}{\partial x^2} = 2 > 0 \), \( (2, 0) \) is a local minimum point.

Key concepts

Gradient Vector

In multivariable calculus, the gradient vector is a crucial concept used to find critical points of a function. It shows the direction of the greatest rate of increase of a function. To calculate the gradient vector of a function like \( f(x, y) = x^2 + 4y^2 - 4x \), we take the partial derivatives with respect to each variable.

• First, partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = 2x - 4 \).
• Then, partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = 8y \).

The gradient vector is: \( abla f = \left( 2x - 4, 8y \right) \). The critical points occur where this gradient vector equals zero.

Solving \( 2x - 4 = 0 \) and \( 8y = 0 \) gives us the critical point \( (2, 0) \). This point can then be analyzed further to determine its nature: whether it's a local maximum, minimum, or a saddle point.

Hessian Matrix

The Hessian matrix is a square matrix of second-order partial derivatives of a scalar-valued function. It provides insights into the local curvature of a function. For our function, the Hessian matrix \( H \) can be expressed as follows:\[H = \begin{pmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} \end{pmatrix}\]For the function \( f(x, y) = x^2 + 4y^2 - 4x \), we calculate:

• \( \frac{\partial^2 f}{\partial x^2} = 2 \)
• \( \frac{\partial^2 f}{\partial y^2} = 8 \)
• \( \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = 0 \)

Thus, the Hessian matrix for our function is:\[H = \begin{pmatrix} 2 & 0 \0 & 8 \end{pmatrix}\] The Hessian matrix helps determine the nature of critical points through its determinant and other characteristics.

Second Partial Derivatives

Second partial derivatives are derivatives of the partial derivatives. They provide further information about the curvature of the function. For example, in our problem, the second derivatives were:

• \( \frac{\partial^2 f}{\partial x^2} = 2 \)
• \( \frac{\partial^2 f}{\partial y^2} = 8 \)
• \( \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = 0 \)

These second partial derivatives form the Hessian matrix. They are used to evaluate the nature of the critical points.

A key property of second partial derivatives is symmetry, which means \( \frac{\partial^2 f}{\partial x \partial y} \) equals \( \frac{\partial^2 f}{\partial y \partial x} \). This characteristic is essential when constructing the Hessian matrix and discussing the function's behavior near critical points.

Local Maximum and Minimum

Finding local maxima and minima involves identifying critical points of a function and using tests like the second derivative test.

The positive determinant of our Hessian matrix \( \det(H) = 16 \) indicates that the method is inconclusive if only considered by itself. However, since \( \frac{\partial^2 f}{\partial x^2} = 2 > 0 \), the critical point \( (2, 0) \) is classified as a local minimum.

• If \( \det(H) \) were negative, it would suggest a saddle point.
• Both negative values of second derivatives would mean a local maximum.

Understanding these rules helps one identify whether a critical point is a local maximum, minimum, or a saddle point. In problems where you have multiple variables, applying the Hessian determinant test is essential in classifying points accurately.