Problem 11 - Calculus with Concepts in Calculus

Question

Find all critical points. Determine whether each critical point yields a relative maximum value, a relative minimum value, or a saddle point. $$ f(x, y)=4 x y+2 x^{2} y-x y^{2} $$

Step-by-Step Solution

Answer

Critical points: (0, 0) and (2, 4); (2, 4) is a local minimum.

Step 1: Calculate Partial Derivatives

First, find the partial derivatives of the function \( f(x, y) = 4xy + 2x^2y - xy^2 \). The partial derivative with respect to \( x \) is \( f_x = \frac{\partial f}{\partial x} = 4y + 4xy - y^2 \). The partial derivative with respect to \( y \) is \( f_y = \frac{\partial f}{\partial y} = 4x + 2x^2 - 2xy \).

Step 2: Set Partial Derivatives to Zero

To find the critical points, set \( f_x = 0 \) and \( f_y = 0 \).\From \( f_x = 0 \): \( 4y + 4xy - y^2 = 0 \).\From \( f_y = 0 \): \( 4x + 2x^2 - 2xy = 0 \).

Step 3: Solve the System of Equations

To solve the equations, we solve: 1. \( y(4 + 4x - y) = 0 \)2. \( 2x(2 + x - y) = 0 \)From these, we have possible cases:a) \( y = 0 \) gives solutions \( x = 0 \).b) \( 4 + 4x - y = 0 \), \( 2 + x - y = 0 \).Solving these, \( x = 0 \) or \( y = 4 + 4x \). For the trunk equation \( 2+x-y = 0 \), align them, you get \( x=2 \), \( y=4 \). So \((2, 4)\) is a critical point. Also solve by substitution to check other possible solutions: creating x=0, y=0 as another critical point.

Step 4: Use the Second Derivative Test

To classify the critical points, calculate the second derivatives: \( f_{xx} = \frac{\partial^2 f}{\partial x^2} = 4y \), \( f_{yy} = \frac{\partial^2 f}{\partial y^2} = -2x \), and \( f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 4 + 4x - 2y \).\Compute the Hessian matrix determinant: \[ D = f_{xx}f_{yy} - (f_{xy})^2 \] for each critical point.

Step 5: Evaluate the Hessian Determinant

At point \((0, 0)\) : \( f_{xx}=0 \), \( f_{yy}=0 \), \( f_{xy} = 4 \). \( D = 0 \) - inconclusive. For (2,4): \(f_{xx} = 16 \), \(f_{yy} = 4 \), \(f_{xy} = 0\). \( D = 64 \) which is positive, and \(f_{xx} > 0\) indicating a local minimum at \((2, 4)\), see if changes made

Key Concepts

Partial Derivatives

To find critical points in a function of two variables, we begin with partial derivatives. Partial derivatives give us the rate of change of the function with respect to one variable while keeping the other constant. For the function \( f(x, y) = 4xy + 2x^2y - xy^2 \):

• The partial derivative with respect to \( x \) is represented as \( f_x = \frac{\partial f}{\partial x} = 4y + 4xy - y^2 \).
• The partial derivative with respect to \( y \) is represented as \( f_y = \frac{\partial f}{\partial y} = 4x + 2x^2 - 2xy \).

These equations help us locate critical points by setting them equal to zero and solving the system to find where the function's slope is zero in both directions.

Second Derivative Test

Once you have the critical points, the Second Derivative Test helps you classify each point as a relative maximum, relative minimum, or a saddle point. This test uses second partial derivatives of the function.
For a function \( f \), the involved second partial derivatives are:

• \( f_{xx} = \frac{\partial^2 f}{\partial x^2} \)
• \( f_{yy} = \frac{\partial^2 f}{\partial y^2} \)
• \( f_{xy} = \frac{\partial^2 f}{\partial x \partial y} \)

The determinant of the Hessian matrix \( D \) is evaluated at the critical point as given by:\[D = f_{xx}f_{yy} - (f_{xy})^2\]Analyzing \( D \) and \( f_{xx} \) lets us decide the nature of each critical point.

Hessian Matrix

The Hessian matrix is a square matrix of second-order partial derivatives of a function. It is crucial in the Second Derivative Test for classifying critical points. The matrix is structured as follows for our function \( f(x, y) = 4xy + 2x^2y - xy^2 \):

• \( f_{xx} = 4y \)
• \( f_{yy} = -2x \)
• \( f_{xy} = f_{yx} = 4 + 4x - 2y \)

Constructing the Hessian gives:\[H = \begin{bmatrix}f_{xx} & f_{xy} \f_{xy} & f_{yy} \\end{bmatrix}\]The determinant of this matrix \( D \) helps determine whether your critical point is a saddle point, minimum, or maximum based on its sign and the sign of \( f_{xx} \).

Relative Extrema

Relative extrema refer to the local minimums and maximums of a function. When applying the Second Derivative Test, after finding \( D \) and checking \( f_{xx} \):

• If \( D > 0 \) and \( f_{xx} > 0 \), there's a local minimum at the point.
• If \( D > 0 \) and \( f_{xx} < 0 \), there's a local maximum.
• If \( D < 0 \), the point is a saddle point (neither a minimum nor maximum).
• If \( D = 0 \), the test is inconclusive.

For instance, in our exercise: at the critical point (2, 4), \( D = 64 \) is positive and \( f_{xx} = 16 \) which is positive, indicating a local minimum. This crucial analysis helps in understanding the functional landscape and optimally using it.