Problem 19 - Calculus Early Transcendentals

Question

Show that \(f(x, y)=x^{2}+4 y^{2}-4 x y+2\) has an infinite number of critical points and that \(D=0\) at each one. Then show that \(f\) has a local (and absolute) minimum at each critical point.

Step-by-Step Solution

Answer

Critical points lie along the line \(x = 2y\) with \(D = 0\), and each is a minimum point.

Step 1: Find the Partial Derivatives

To find the critical points of the function \(f(x, y) = x^2 + 4y^2 - 4xy + 2\), we need to calculate its partial derivatives with respect to \(x\) and \(y\). The partial derivative with respect to \(x\) is \(f_x = \frac{\partial}{\partial x}(x^2 + 4y^2 - 4xy + 2) = 2x - 4y\). The partial derivative with respect to \(y\) is \(f_y = \frac{\partial}{\partial y}(x^2 + 4y^2 - 4xy + 2) = 8y - 4x\).

Step 2: Set Partial Derivatives to Zero

Set the partial derivatives equal to zero to find the critical points: \[ 2x - 4y = 0 \] \[ 8y - 4x = 0 \]From \(2x - 4y = 0\), we get \(x = 2y\). Substituting \(x = 2y\) into the second equation \(8y - 4x = 0\) gives \(8y - 4(2y) = 0\), which is true for any \(y\). Therefore, critical points occur along the line \(x = 2y\).

Step 3: Evaluate the Hessian Determinant

We need to calculate the second partial derivatives to evaluate the Hessian determinant:\(f_{xx} = \frac{\partial^2}{\partial x^2}(f) = 2\), \(f_{yy} = \frac{\partial^2}{\partial y^2}(f) = 8\), \(f_{xy} = \frac{\partial^2}{\partial x \partial y}(f) = -4\).The Hessian determinant is given by: \[ D = f_{xx}f_{yy} - (f_{xy})^2 = 2 \times 8 - (-4)^2 = 16 - 16 = 0 \]

Step 4: Analyze the Nature of Critical Points

Since the Hessian determinant \(D = 0\), the second derivative test is inconclusive. However, any line equation like \(x = 2y\) that satisfies the partial derivatives represents a trough, and as seen from the expression \(f(x, y) = (x - 2y)^2 + 2\), it's always non-negative and is zero only when \(x = 2y\). Hence for all critical points along this line, \(f\) achieves its minimum value \(2\).

Step 5: Conclusion

The function \(f(x, y)=x^{2}+4y^{2}-4xy+2\) indeed has an infinite number of critical points along the line where \(x = 2y\), and \(D = 0\) for each of these points. Each point is a minimum point with minimum value \(2\).

Key Concepts

Partial Derivatives

In multivariable calculus, the concept of partial derivatives helps us understand how a function changes as we vary only one of the variables. For a function like \(f(x, y)=x^2+4y^2-4xy+2\), calculating the partial derivatives with respect to \(x\) and \(y\) gives us the gradients in each direction individually.

- The partial derivative with respect to \(x\), denoted as \(f_x\), shows how the function changes as \(x\) changes, holding \(y\) constant.- Similarly, the partial derivative with respect to \(y\), denoted as \(f_y\), reveals how the function varies as \(y\) changes, holding \(x\) constant.For our function, the calculations are:

• \(f_x = 2x - 4y\)
• \(f_y = 8y - 4x\)

These derivatives point us to the critical points, which are places where the slope is zero, meaning a potential maximum, minimum, or saddle point. To find these, we set \(f_x = 0\) and \(f_y = 0\), leading us to the line \(x = 2y\), indicating an infinite number of critical points along this line.

Hessian Determinant

The Hessian matrix is employed to investigate the nature of critical points by examining second derivatives. For a function of two variables, the Hessian matrix is\[H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{bmatrix}\]The determinant of this matrix, known as the Hessian determinant, is a crucial factor. It is calculated as:\[ D = f_{xx}f_{yy} - (f_{xy})^2 \]For our function \(f(x, y)\), we computed:

• \(f_{xx} = 2\)
• \(f_{yy} = 8\)
• \(f_{xy} = -4\)

Substituting these into the formula for \(D\) gives us:\[ D = 2 \times 8 - (-4)^2 = 16 - 16 = 0 \]This zero result implies that the second derivative test does not provide us with conclusive information about whether the critical points are maxima, minima, or saddle points.

Second Derivative Test

The second derivative test involves using the Hessian determinant to classify critical points. Generally:

• If \(D > 0\) and \(f_{xx} > 0\), the function has a local minimum at the critical point.
• If \(D > 0\) and \(f_{xx} < 0\), the function has a local maximum.
• If \(D < 0\), the function has a saddle point.
• When \(D = 0\), as in our case, the test is inconclusive.

Since \(D = 0\) for the function \(f(x, y) = x^2 + 4y^2 - 4xy + 2\), we cannot directly determine the nature of the critical points using this test. However, upon inspecting further, specifically analyzing the structure \((x - 2y)^2 + 2\) of our function, we see that these points represent a definite trough or valley, indicating a minimum.

Local and Absolute Minimum

A local minimum is a point where the function value is lower than all points in a small surrounding area. An absolute minimum is lower than or equal to function values at all other points in the entire domain. In the given function \(f(x, y)=x^2 + 4y^2 - 4xy + 2\), each critical point occurs along the line \(x = 2y\).

By rewriting the function as \((x - 2y)^2 + 2\), we see the expression is always non-negative, meaning the minimum value happens where \((x - 2y)^2 = 0\). This occurs precisely at the line \(x = 2y\), maintaining consistency at every critical point.

Thus, every critical point on the line \(x = 2y\) serves as a local minimum. Not only that, but since the entire expression resolves to the constant \(2\) at those points, it is also the function's absolute minimum. This uniform minimum further confirms our analysis of critical points being minimal along this line.