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Locate and classify any critical points. $$ g(x, y)=4 x y-x^{4}-y^{4} $$

Short Answer

Expert verified
Critical points are (0,0) (saddle), (1,1) (local max), (-1,-1) (local max).

Step by step solution

01

Find the First Partial Derivatives

To locate critical points of the function \( g(x, y) = 4xy - x^4 - y^4 \), we first need to find the first partial derivatives with respect to \( x \) and \( y \). The partial derivative with respect to \( x \) is \( g_x(x, y) = 4y - 4x^3 \). The partial derivative with respect to \( y \) is \( g_y(x, y) = 4x - 4y^3 \).
02

Set the First Partial Derivatives to Zero

Next, we set the first partial derivatives equal to zero to find the critical points: \( 4y - 4x^3 = 0 \) and \( 4x - 4y^3 = 0 \). Simplifying these equations gives us \( y = x^3 \) and \( x = y^3 \).
03

Solve the System of Equations

Substitute \( y = x^3 \) into \( x = y^3 \), getting \( x = (x^3)^3 \) which simplifies to \( x = x^9 \). Solving \( x - x^9 = 0 \), we factor to find \( x(1 - x^8) = 0 \). This gives \( x = 0 \) or \( x^8 = 1 \), leading to \( x = 1 \) and \( x = -1 \) as additional solutions.
04

Find Corresponding y-values

For each solution for \( x \), substitute back to find \( y \). When \( x = 0 \), \( y = 0^3 = 0 \). When \( x = 1 \), \( y = 1^3 = 1 \). When \( x = -1 \), \( y = (-1)^3 = -1 \). Accordingly, the critical points are \((0,0)\), \((1,1)\), and \((-1,-1)\).
05

Classify the Critical Points Using the Second Derivative Test

To classify the critical points, compute the second derivatives: \( g_{xx} = -12x^2 \), \( g_{yy} = -12y^2 \), \( g_{xy} = 4 \). Calculate the Hessian determinant \( H = g_{xx}g_{yy} - (g_{xy})^2 \). For \( g(x,y) \), \( H = (-12x^2)(-12y^2) - 16 \). Evaluate at each critical point to classify: \( (0,0) \) gives \( H = -16 \) (saddle), \((1,1)\) gives \( H = 128 \) (local max), and \((-1,-1)\) gives \( H = 128 \) (local max).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
In multivariable calculus, partial derivatives are essential for finding critical points of a function. For the function \( g(x, y) = 4xy - x^4 - y^4 \), partial derivatives reveal how the function changes with respect to one variable while keeping the other constant. To find these, we differentiate the function with respect to each variable separately.
\( g_x(x, y) \) represents the rate of change of the function as \( x \) varies, which simplifies to \( 4y - 4x^3 \). Similarly, \( g_y(x, y) \) represents the rate of change as \( y \) changes, simplifying to \( 4x - 4y^3 \).
By setting these partial derivatives to zero, \( g_x(x, y) = 0 \) and \( g_y(x, y) = 0 \), we find critical points where the function may have a maximum, minimum, or saddle point.
  • Partial Derivative with respect to \( x \): \( 4y - 4x^3 \)
  • Partial Derivative with respect to \( y \): \( 4x - 4y^3 \)
Second Derivative Test
Once we have located potential critical points, the second derivative test helps us classify them. In this test, we examine the second partial derivatives of the function. For \( g(x, y) \), the second partial derivatives are \( g_{xx} = -12x^2 \) and \( g_{yy} = -12y^2 \).
The second derivative test involves the Hessian determinant, which we will discuss shortly. But fundamentally, this test indicates whether a critical point is a local maximum, local minimum, or saddle point by analyzing the concavity of the function around those points. The behavior surrounding the critical points tells us which type each point is.
  • Second Partial Derivative with respect to \( x \): \( g_{xx} = -12x^2 \)
  • Second Partial Derivative with respect to \( y \): \( g_{yy} = -12y^2 \)
Hessian Determinant
The Hessian determinant, denoted as \( H \), is crucial for the second derivative test. It's constructed from the second partial derivatives:
\[ H = g_{xx}g_{yy} - (g_{xy})^2 \]
Here, \( g_{xy} \) is the mixed partial derivative, which for this function is 4. The Hessian tells us about the nature of a critical point:
- If \( H > 0 \) and \( g_{xx} > 0 \), we have a local minimum.- If \( H > 0 \) and \( g_{xx} < 0 \), it's a local maximum.- If \( H < 0 \), the critical point is a saddle point.- If \( H = 0 \), the test is inconclusive.Calculating the Hessian at each critical point gives us clarity on their types.
Classification of Critical Points
Finally, classifying critical points is the key outcome of our computations. We apply the second derivative test and the Hessian determinant to each critical point to determine its nature.
For \((x, y)\):
  • At \((0,0)\), \( H = -16 \), indicating a saddle point.
  • At \((1,1)\), \( H = 128 \), suggesting a local maximum because \( g_{xx} < 0 \).
  • At \((-1,-1)\), also \( H = 128 \), again showing a local maximum.
By identifying these classifications, we deepen our understanding of the function's geometrical nature, showing where it rises, falls, or changes direction.

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