Questions: Apply a second derivative to identify a critical points as a local maximum, local minimum or saddle point for a function. Suppose that f(x, y)=2 x^4+2 y^4-1 x y then the minimum is

Solution Steps

To identify the critical points and determine whether they are local maxima, local minima, or saddle points, we need to follow these steps:

1. Compute the first partial derivatives of the function \( f(x, y) \).
2. Set the first partial derivatives equal to zero to find the critical points.
3. Compute the second partial derivatives to form the Hessian matrix.
4. Evaluate the Hessian determinant and the second partial derivatives at the critical points to classify them.

To find the critical points, we first need to compute the first partial derivatives of the function \( f(x, y) = 2x^4 + 2y^4 - xy \).

\[ f_x = \frac{\partial}{\partial x}(2x^4 + 2y^4 - xy) = 8x^3 - y \]

\[ f_y = \frac{\partial}{\partial y}(2x^4 + 2y^4 - xy) = 8y^3 - x \]

To find the critical points, we set the first partial derivatives equal to zero:

\[ 8x^3 - y = 0 \quad \text{(1)} \]

\[ 8y^3 - x = 0 \quad \text{(2)} \]

From equation (1):

\[ y = 8x^3 \]

Substitute \( y = 8x^3 \) into equation (2):

\[ 8(8x^3)^3 - x = 0 \]

\[ 8 \cdot 512x^9 - x = 0 \]

\[ 4096x^9 - x = 0 \]

Factor out \( x \):

\[ x(4096x^8 - 1) = 0 \]

This gives us two solutions:

\[ x = 0 \quad \text{or} \quad 4096x^8 = 1 \]

For \( 4096x^8 = 1 \):

\[ x^8 = \frac{1}{4096} \]

\[ x = \pm \left(\frac{1}{4096}\right)^{\frac{1}{8}} = \pm \frac{1}{2} \]

So, the critical points are:

\[ (x, y) = (0, 0), \left(\frac{1}{2}, 1\right), \left(-\frac{1}{2}, -1\right) \]

Next, we compute the second partial derivatives:

\[ f_{xx} = \frac{\partial}{\partial x}(8x^3 - y) = 24x^2 \]

\[ f_{yy} = \frac{\partial}{\partial y}(8y^3 - x) = 24y^2 \]

\[ f_{xy} = \frac{\partial}{\partial y}(8x^3 - y) = -1 \]

\[ f_{yx} = \frac{\partial}{\partial x}(8y^3 - x) = -1 \]

The Hessian matrix \( H \) is:

\[ H = \begin{pmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{pmatrix} = \begin{pmatrix} 24x^2 & -1 \\ -1 & 24y^2 \end{pmatrix} \]

The determinant of the Hessian matrix is:

\[ \text{det}(H) = (24x^2)(24y^2) - (-1)(-1) = 576x^2y^2 - 1 \]

1. At \( (0, 0) \):

\[ \text{det}(H) = 576(0)^2(0)^2 - 1 = -1 \]

Since the determinant is negative, \( (0, 0) \) is a saddle point.

2. At \( \left(\frac{1}{2}, 1\right) \):

\[ \text{det}(H) = 576\left(\frac{1}{2}\right)^2(1)^2 - 1 = 576 \cdot \frac{1}{4} - 1 = 144 - 1 = 143 \]

Since the determinant is positive and \( f_{xx} > 0 \):

\[ f_{xx} = 24\left(\frac{1}{2}\right)^2 = 6 > 0 \]

Thus, \( \left(\frac{1}{2}, 1\right) \) is a local minimum.

3. At \( \left(-\frac{1}{2}, -1\right) \):

\[ \text{det}(H) = 576\left(-\frac{1}{2}\right)^2(-1)^2 - 1 = 576 \cdot \frac{1}{4} - 1 = 144 - 1 = 143 \]

Since the determinant is positive and \( f_{xx} > 0 \):

\[ f_{xx} = 24\left(-\frac{1}{2}\right)^2 = 6 > 0 \]

Thus, \( \left(-\frac{1}{2}, -1\right) \) is a local minimum.

Final Answer