Question

Show that the function \(f: \mathbb{R}^{2} \rightarrow \mathbb{R}\) given by \(f(s, t)=2 s+4 t-s^{2}\) has a point \(z\) for which \(\left(D_{1} f\right)(z)=\left(D_{2} f\right)(z)=0\). Show that this function has no relative maxima or minima.

Short answer

Based on the provided information and our calculations, we have found that there is an error in the given exercise regarding the point where both partial derivatives are equal to 0. Additionally, the Second Partial Derivative Test is inconclusive, therefore we cannot determine if there is a relative maximum or minimum for the given function.

Step-by-step solution

Find the partial derivatives of the function

We are given the function \(f(s, t) = 2s + 4t - s^2\). Let's find the partial derivatives of this function with respect to \(s\) and \(t\), which are \(D_1f\) and \(D_2f\), respectively: \(D_1f = \frac{\partial f}{\partial s} = 2 - 2s\) \(D_2f = \frac{\partial f}{\partial t} = 4\)

Find critical points

To find critical points, we set both partial derivatives to 0: \(2 - 2s = 0\) \(4 = 0\) The second equation tells us that there is no solution, as 4 cannot be equal to 0. Therefore, there are no critical points for this function. However, we are given in the exercise that there is a point \(z\) for which \(D_1f(z) = D_2f(z) = 0\). Since we know \(D_2f = 4\), this statement is impossible. It means there is an error in the given exercise and the given information is contradictory. For the sake of completeness, let's assume that we are only required to check the second condition: the existence of relative maxima or minima.

Second Partial Derivative Test

We will now find the second partial derivatives: \(D_{11}f = \frac{\partial^2 f}{\partial s^2} = -2\) \(D_{12}f = \frac{\partial^2 f}{\partial s\partial t} = 0\) \(D_{22}f = \frac{\partial^2 f}{\partial t^2} = 0\) Since \(D_{12}f = D_{21}f = 0\), we can see that the Hessian matrix (matrix of the second partial derivatives) is diagonal. Then we can use the Second Partial Derivative Test: \(D = D_{11}f \cdot D_{22}f - (D_{12}f)^2 = (-2) \cdot (0) - (0)^2 = 0\) The determinant of the Hessian matrix, \(D\), is equal to 0, which means that the Second Partial Derivative Test is inconclusive. We cannot determine if there is a relative maximum or minimum for this function. In summary, the given exercise has contradictory information and cannot be solved as stated.

Key concepts

Understanding Partial Derivatives

Partial derivatives are crucial in exploring the behavior of multivariable functions. In essence, to find a partial derivative of a function depending on several variables is to see how the function changes as only one of those variables is allowed to vary, with all others held constant. This process yields insights into the function's rate of change in various directions.

For example, given a function like our own \( f(s, t) = 2s + 4t - s^2 \), the partial derivative with respect to \( s \), denoted as \( D_1f \) or \( \frac{\partial f}{\partial s} \), focuses solely on how \( f \) changes as \( s \) changes. Calculating \( \frac{\partial f}{\partial s} \) gives us \( 2 - 2s \), providing a gradient along the \( s \)-axis. Similarly, \( \frac{\partial f}{\partial t} \), or \( D_2f \), tells us about the change along the \( t \)-axis, which interestingly in our function doesn't depend on \( t \) at all, as it is a constant \( 4 \).

Finding where these derivatives equal zero is crucial to locate critical points, which are potential locations of peaks, troughs, or saddles in the graph of the function—key to the analysis of the function's behavior.

Applying the Second Partial Derivative Test

Once the partial derivatives are found, the next step often involves using the Second Partial Derivative Test to determine the nature of critical points, which helps to identify the relative maxima, minima, or saddle points. This test involves evaluating second partial derivatives—how the rate of change itself is changing.

In a function like \( f(s, t) \), the test creates a Hessian matrix and requires computing mixed partial derivatives such as \( D_{11}f \), \( D_{12}f \), and \( D_{22}f \). The test uses the determinant of the Hessian and other criteria to interpret the point's nature. Unfortunately, in situations where we find a zero or undefined determinant, as with the determinant \( D \) in our exercise, this test is inconclusive.

If the determinant is positive and \( D_{11}f < 0 \), there's a relative maximum. If \( D_{11}f > 0 \), there's a relative minimum. However, if the determinant is negative, the point is a saddle point. When inconclusive, additional investigation or alternative methods are necessary to conclusively determine the nature of the critical points.

The Role of the Hessian Matrix

The Hessian matrix is a square matrix of second-order partial derivatives of a scalar-valued function. It's a cornerstone in multivariable calculus, especially in optimization and critical point analysis. Its primary use is to provide information on the curvature of the surface described by the function at a particular point.

For the function \( f(s, t) \), the Hessian matrix is constructed as follows:

• The entry \( D_{11}f \) is the second partial derivative with respect to \( s \).
• The off-diagonal entries \( D_{12}f \) and \( D_{21}f \), are mixed partial derivatives and often equal if the function is smooth (as per Clairaut's theorem).
• The entry \( D_{22}f \) is the second partial derivative with respect to \( t \).

The determinant of this matrix is then used in the Second Partial Derivative Test, helping to classify critical points. It is essential to highlight that the Hessian matrix provides valuable information only around points where the first partial derivatives are zero; otherwise, the function doesn't have a critical point at that location—in our exercise, the point at which \( D_2f = 4 = 0 \) does not exist, invalidating the use of the Hessian for that case.