Problem 27
Question
Find the critical points and test for relative extrema. List the critical points for which the Second-Partials Test fails. $$ f(x, y)=x^{3}+y^{3} $$
Step-by-Step Solution
Verified Answer
The only critical point of the function \(f(x, y) = x^{3} + y^{3}\) is at the origin (0,0). However, the Second Partials Test fails at this point, so it's not possible to determine whether this point is a relative maximum, a relative minimum, or a saddle point based on this method alone.
1Step 1: Compute the partial derivatives
Start by calculating the first-order partial derivatives of the given function. They are given by \(\frac{\partial f}{\partial x} = 3x^{2}\) and \(\frac{\partial f}{\partial y} = 3y^{2}\).
2Step 2: Find critical points
The critical points occur where the partial derivatives are both 0 or undefined. Here they are both defined for all inputs, and they are only 0 when x = 0 and y = 0. Therefore the only critical point of f is (0, 0).
3Step 3: Apply the Second Partials Test
Now, we apply the Second-Partials Test to classify the critical point. First, find the second-order derivatives: \(\frac{\partial^{2} f}{\partial x^{2}} = 6x\), \(\frac{\partial^{2} f}{\partial y^{2}} = 6y\), \(\frac{\partial^{2} f}{\partial x\partial y} = 0\). Then, compute the determinant of the Hessian matrix at (0,0): \(D = \frac{\partial^{2} f}{\partial x^{2}} * \frac{\partial^{2} f}{\partial y^{2}} - \left(\frac{\partial^{2} f}{\partial x\partial y}\right)^{2} = 6x * 6y\). At the point (0,0), D = 0 which means that the Second Partials Test fails.
4Step 4: Conclusion
Since the Second Partials Test is inconclusive, we can't confirm whether the point (0,0) is a relative maximum, a relative minimum, or a saddle point. Extra methods would have to be employed to classify this critical point.
Key Concepts
Critical Points in CalculusPartial DerivativesHessian Matrix
Critical Points in Calculus
In calculus, critical points play a crucial role in understanding the behavior of functions, especially when analyzing their graphs. Critical points are locations on a graph where the function's derivative is zero or undefined. These points signify the potential locations of relative maxima, minima, or saddle points.
In the context of multivariable functions, such as the function \( f(x, y) = x^{3} + y^{3} \) from our exercise, critical points are where the partial derivatives of the function equal zero or don't exist. To find these critical points, one must calculate the first-order partial derivatives with respect to each variable and solve for when these derivatives are zero. The outcomes of this calculation, as demonstrated in the exercise, yield the coordinates of critical points. In the given example, the critical point is at (0, 0).
Understanding critical points is essential as they are often used to optimize functions — for instance, in economics to find maximum profit or in engineering to minimize material stress.
In the context of multivariable functions, such as the function \( f(x, y) = x^{3} + y^{3} \) from our exercise, critical points are where the partial derivatives of the function equal zero or don't exist. To find these critical points, one must calculate the first-order partial derivatives with respect to each variable and solve for when these derivatives are zero. The outcomes of this calculation, as demonstrated in the exercise, yield the coordinates of critical points. In the given example, the critical point is at (0, 0).
Understanding critical points is essential as they are often used to optimize functions — for instance, in economics to find maximum profit or in engineering to minimize material stress.
Partial Derivatives
Partial derivatives are a fundamental concept when dealing with functions of several variables. While a derivative of a single-variable function informs us about the rate at which the function values change with respect to changes in the input, partial derivatives give us similar information for multivariable functions.
A partial derivative, represented as \( \frac{\partial f}{\partial x} \), measures how the function changes as one specific variable changes, holding all other variables constant. This allows us to understand the function's behavior in the direction of each individual variable.
In our exercise involving the function \( f(x, y) = x^{3} + y^{3} \), the first-order partial derivatives are \( 3x^{2} \) and \( 3y^{2} \) with respect to \( x \) and \( y \), respectively. These equations are set to zero to find the critical points, which shows that the impact of each individual variable on the function's output is neutral at the critical point.
A partial derivative, represented as \( \frac{\partial f}{\partial x} \), measures how the function changes as one specific variable changes, holding all other variables constant. This allows us to understand the function's behavior in the direction of each individual variable.
In our exercise involving the function \( f(x, y) = x^{3} + y^{3} \), the first-order partial derivatives are \( 3x^{2} \) and \( 3y^{2} \) with respect to \( x \) and \( y \), respectively. These equations are set to zero to find the critical points, which shows that the impact of each individual variable on the function's output is neutral at the critical point.
Hessian Matrix
The Hessian matrix is a square matrix of second-order partial derivatives of a multivariable function. Crucially, it is used in the Second-Partials Test to determine the nature of critical points — that is, whether such points are relative maxima, minima, or saddle points. Named after the German mathematician Ludwig Otto Hesse, this matrix encapsulates the curvature information of the function's graph at a given point.
The Hessian matrix for a two-variable function \( f(x, y) \) is a 2x2 matrix containing all possible mixed second derivatives. The determinant of the Hessian matrix at a critical point, often denoted as \( D \), is the key factor for the Second-Partials Test. For the function in our exercise, the Hessian matrix at (0, 0) has a determinant of zero, indicating that the test fails to provide a definitive answer about the nature of the critical point.
Mathematically, the Hessian matrix for our example function at the critical point (0, 0) would be a matrix with zeros in all positions because the mixed partial derivative is zero and the second order partial derivatives become zero when evaluated at (0, 0). This outcome requires further exploration with additional methods to classify the critical point.
The Hessian matrix for a two-variable function \( f(x, y) \) is a 2x2 matrix containing all possible mixed second derivatives. The determinant of the Hessian matrix at a critical point, often denoted as \( D \), is the key factor for the Second-Partials Test. For the function in our exercise, the Hessian matrix at (0, 0) has a determinant of zero, indicating that the test fails to provide a definitive answer about the nature of the critical point.
Mathematically, the Hessian matrix for our example function at the critical point (0, 0) would be a matrix with zeros in all positions because the mixed partial derivative is zero and the second order partial derivatives become zero when evaluated at (0, 0). This outcome requires further exploration with additional methods to classify the critical point.
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