Problem 12 - Applied Calculus

Question

Find all the critical points and determine whether each is a local maximum, local minimum, a saddle point, or none of these. $$f(x, y)=x^{3}+y^{3}-3 x^{2}-3 y+10$$

Step-by-Step Solution

Answer

Critical points are (0, 1) saddle, (0, -1) local max, (2, 1) local min, (2, -1) saddle.

Step 1: Find the Partial Derivatives

To find the critical points, we need the first partial derivatives of the function.The partial derivative with respect to \(x\) is:\[ f_x = \frac{\partial}{\partial x}(x^3 + y^3 - 3x^2 - 3y + 10) = 3x^2 - 6x \]The partial derivative with respect to \(y\) is:\[ f_y = \frac{\partial}{\partial y}(x^3 + y^3 - 3x^2 - 3y + 10) = 3y^2 - 3 \]

Step 2: Solve for Critical Points

Set the first partial derivatives equal to zero and solve for \(x\) and \(y\).For \(f_x = 0\):\[ 3x^2 - 6x = 0 \]Simplify this to:\[ 3x(x - 2) = 0 \]Giving solutions \(x = 0\) and \(x = 2\).For \(f_y = 0\):\[ 3y^2 - 3 = 0 \]Simplify this to:\[ 3(y^2 - 1) = 0 \]Giving solutions \(y = 1\) and \(y = -1\).Thus, the potential critical points are \((0, 1), (0, -1), (2, 1), (2, -1)\).

Step 3: Use the Second Derivative Test

Calculate the second partial derivatives:\(f_{xx} = \frac{\partial^2 f}{\partial x^2} = 6x - 6\),\(f_{yy} = \frac{\partial^2 f}{\partial y^2} = 6y\),\(f_{xy} = f_{yx} = \frac{\partial^2 f}{\partial x \partial y} = 0\).The Hessian determinant is:\[ H = f_{xx}f_{yy} - (f_{xy})^2 = (6x - 6)(6y) - 0 \].

Step 4: Evaluate the Hessian at Critical Points

Evaluate the Hessian at each critical point to determine the nature of each point:1. For \((0, 1)\): \(f_{xx} = -6\), \(f_{yy} = 6\), and \(H = (-6)(6) = -36 < 0\). This is a saddle point.2. For \((0, -1)\): \(f_{xx} = -6\), \(f_{yy} = -6\), and \(H = (-6)(-6) = 36 > 0\), with \(f_{xx} < 0\), indicating a local maximum.3. For \((2, 1)\): \(f_{xx} = 6\), \(f_{yy} = 6\), and \(H = (6)(6) = 36 > 0\), with \(f_{xx} > 0\), indicating a local minimum.4. For \((2, -1)\): \(f_{xx} = 6\), \(f_{yy} = -6\), and \(H = (6)(-6) = -36 < 0\). This is a saddle point.

Key Concepts

Partial Derivatives

Partial derivatives are fundamental in calculating critical points for multivariable functions. They provide information on how the function changes with respect to one variable, while keeping the other variable constant. In this exercise, two partial derivatives were calculated:

• The partial derivative with respect to x, denoted as \( f_x \), is computed by treating \( y \) as a constant and differentiating the function with respect to \( x \). Here, \( f_x = 3x^2 - 6x \).
• The partial derivative with respect to y, denoted as \( f_y \), treats \( x \) as a constant, resulting in \( f_y = 3y^2 - 3 \).

By setting these partial derivatives to zero, we can determine the stationary points of the function, where potential maximums, minimums, or saddle points may occur.

Second Derivative Test

The second derivative test extends the idea of the second derivative in single-variable calculus to functions of multiple variables. This test is crucial in determining the nature of critical points. After finding the first partial derivatives and solving them to obtain the critical points, the second partial derivatives are needed. In this exercise, three second derivatives were calculated:

• \( f_{xx} \), the second derivative with respect to \( x \), is given by \( 6x - 6 \).
• \( f_{yy} \), the second derivative with respect to \( y \), is \( 6y \).
• \( f_{xy} = f_{yx} \), the mixed derivative, is zero for this function.

These derivatives help compute the Hessian determinant, which is essential for the second derivative test.

Hessian Determinant

The Hessian determinant uses second derivatives to decide what kind of critical point is present. It is calculated as: \[ H = f_{xx}f_{yy} - (f_{xy})^2 \] For this exercise, the mixed partial derivatives \( f_{xy} \) and \( f_{yx} \) are zero, simplifying the calculation. The determinant becomes \( H = (6x - 6)(6y) \). The value of \( H \) at each critical point helps classify the critical point into one of three categories: local maximum, local minimum, or saddle point, depending on its sign and the sign of \( f_{xx} \).

Local Maximum

A local maximum occurs at a point where the function's value is higher than all nearby points. In the context of the second derivative test, if the Hessian determinant \( H > 0 \) and \( f_{xx} < 0 \), the point is a local maximum. In this exercise, the point \((0, -1)\) was evaluated, where:

• \( f_{xx} = -6 \)
• \( f_{yy} = -6 \)
• The Hessian determinant \( H = 36 > 0 \)

These results satisfy the criteria for a local maximum, indicating that \((0, -1)\) is indeed a point where the function reaches a higher local value compared to its surroundings.

Local Minimum

A local minimum is a point where the function's value is lower than that of all immediate surrounding points. According to the second derivative test, if \( H > 0 \) and \( f_{xx} > 0 \), it suggests a local minimum. In the current problem, the point \((2, 1)\) was considered. At this point:

• \( f_{xx} = 6 \)
• \( f_{yy} = 6 \)
• The Hessian determinant \( H = 36 > 0 \)

These conditions confirm the presence of a local minimum, indicating that \((2, 1)\) is a point where the function falls to its smallest local value.

Saddle Point

A saddle point is a more complex type of critical point, where the surface curves upwards in one direction and downwards in another. In terms of the Hessian determinant, a saddle point is identified when \( H < 0 \). This means there is both a rising and a falling path through that point. For the function explored, two saddle points were identified at:

• \((0, 1)\) with \( H = -36 < 0 \)
• \((2, -1)\) with \( H = -36 < 0 \)

This negative Hessian value indicates that these critical points do not represent a local maxima or minima, but rather a transition point where the curvature changes direction.