Question

Find the stationary values of $$ f(x, y)=4 x^{2}+4 y^{2}+x^{4}-6 x^{2} y^{2}+y^{4} $$ and classify them as maxima, minima or saddle points. Make a rough sketch of the contours of \(f\) in the quarter plane \(x, y \geq 0\).

Short answer

The stationary point at \((0, 0)\) is a local minimum. A rough contour sketch shows elliptical contours centered at the origin.

Step-by-step solution

Find the First Partial Derivatives

To find the stationary points, first calculate the first partial derivatives of the function with respect to both variables. Let's denote the function as \( f(x, y) = 4x^2 + 4y^2 + x^4 - 6x^2y^2 + y^4 \).The first partial derivatives are:\( f_x = \frac{\partial f}{\partial x} = 8x + 4x^3 - 12xy^2 \) and \( f_y = \frac{\partial f}{\partial y} = 8y + 4y^3 - 12x^2y \).

Set the Partial Derivatives to Zero

Set both first partial derivatives to zero and solve for \(x\) and \( y \):\( 8x + 4x^3 - 12xy^2 = 0 \) and \( 8y + 4y^3 - 12x^2y = 0 \).

Solve for \(x\) and \(y\)

Factor each equation to solve for the variables. Starting with \( f_x = 0 \):\( 4x(2 + x^2 - 3y^2) = 0 \). So, \( x = 0 \) or \( 2 + x^2 - 3y^2 = 0 \).For \( f_y = 0 \):\( 4y(2 + y^2 - 3x^2) = 0 \). So, \( y = 0 \) or \( 2 + y^2 - 3x^2 = 0 \).

Find Possible Solutions

From the equations, we find the values of \(x\) and \(y\):When \( x = 0 \): \(8y + 4y^3 = 0\) simplifies to \( y(2 + y^2) = 0 \), giving \( y = 0 \).When \( y = 0 \): \(8x + 4x^3 = 0\) simplifies to \( x(2 + x^3) = 0 \), giving \( x = 0 \). Thus, one stationary point is \((0, 0)\). Consider other combinations of \( x \) and \( y \) from the solutions above.

Evaluate Second Partials for Classification

Use the second partial derivatives to classify the stationary point(s). The second partial derivatives are:\( f_{xx} = 8 + 12x^2 - 12y^2 \)\( f_{yy} = 8 + 12y^2 - 12x^2 \)\( f_{xy} = -24xy \). Evaluate at \((0, 0)\):\( f_{xx}(0, 0) = 8 \)\( f_{yy}(0, 0) = 8 \)\( f_{xy}(0, 0) = 0 \). Calculate the determinant of the Hessian matrix: \( D = f_{xx}f_{yy} - (f_{xy})^2 = 8 \times 8 - 0 = 64 \). As \( D > 0 \) and \( f_{xx} > 0 \), \((0, 0)\) is a local minimum.

Sketch the Contours in the Quarter Plane

Plot the function \(f(x, y)\) in the quadrant where \( x, y \geq 0 \). Sketch level curves by solving \(4x^2 + 4y^2 + x^4 - 6x^2y^2 + y^4 = C\) for various values of \(C\). Note the nature of the contours around the critical point \((0, 0)\) to show it as a minimum.

Key concepts

Partial Derivatives

Partial derivatives are used to understand how a function changes with respect to each variable independently. For a function of two variables, say \( f(x, y) \), the partial derivative with respect to \( x \), denoted as \( f_x \) or \( \frac{\partial f}{\partial x} \), measures how \( f \) changes when only \( x \) varies, keeping \( y \) constant. Similarly, \( f_y \) is the partial derivative with respect to \( y \).

In the given exercise, we found the first partial derivatives of the function \( f(x, y) = 4x^2 + 4y^2 + x^4 - 6x^2y^2 + y^4 \). By setting these partial derivatives to zero, we identify the stationary points, which are potential candidates for maxima, minima, or saddle points.

Hessian Matrix

The Hessian matrix is a square matrix of second-order partial derivatives of a scalar-valued function. For a function \( f(x, y) \), the Hessian matrix \( H \) is:

\[ H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{bmatrix} \]

Here, \( f_{xx} \), \( f_{yy} \) are the second partial derivatives with respect to \( x \) and \( y \), respectively, and \( f_{xy} \) is the mixed partial derivative.

In analyzing the stationary points, the Hessian matrix helps determine their nature. By evaluating the determinants and signs, one can classify the points as local maxima, minima, or saddle points. For our example, we calculated the second partial derivatives and assembled the Hessian matrix at the stationary point \( (0, 0) \), allowing us to classify this point as a local minimum.

Local Minimum

A local minimum of a function occurs at a point where the function value is lower than at nearby points. At a local minimum, small perturbations around this point will not decrease the function's value any further. To classify a stationary point as a local minimum, the determinants of the Hessian matrix's conditions must be met:

- The determinant \( D = f_{xx}f_{yy} - (f_{xy})^2 \) must be greater than zero.
- The second partial derivative \( f_{xx} \) must be positive.

In the provided problem, after calculating the Hessian matrix's determinant and the second-order partial derivatives at the point \( (0, 0) \), we determined \( D > 0 \) and \( f_{xx} > 0 \), confirming that (0, 0) is a local minimum.

Contour Plotting

Contour plotting is a technique used to visualize the level curves of a function of two variables. A contour plot maps the curves on which the function takes on same value, known as contour lines. This graphical representation helps us understand the function's behavior and the nature of its critical points.

To sketch the contours for the given function, we solve \( 4x^2 + 4y^2 + x^4 - 6x^2y^2 + y^4 = C \), where \( C \) is a constant, and plot these curves in the quarter plane where \( x, y \geq 0 \).

We observed that around the stationary point \( (0, 0) \), our contour lines suggest it is a local minimum, coherent with our earlier classification using the Hessian matrix.