Problem 14 - Applied Calculus

Question

Find all the critical points and determine whether each is a local maximum, local minimum, or neither. $$ f(x, y)=x^{3}+y^{3}-3 x^{2}-3 y+10 $$

Step-by-Step Solution

Answer

Critical points: (0,1) and (2,-1) are saddle points, (0,-1) is a local maximum, (2,1) is a local minimum.

Step 1: Find the first partial derivatives

To find critical points of the function \(f(x, y) = x^3 + y^3 - 3x^2 - 3y + 10\), first calculate the partial derivatives with respect to \(x\) and \(y\).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

To find the critical points, set the first partial derivatives equal to zero and solve the system of equations.For \(f_x = 0\):\[ 3x^2 - 6x = 0 \]Factor out the common term:\[ 3x(x - 2) = 0 \]This gives: \(x = 0\) or \(x = 2\).For \(f_y = 0\):\[ 3y^2 - 3 = 0 \]Simplify and solve:\[ y^2 = 1 \]Thus, \(y = 1\) or \(y = -1\).The potential critical points are: \((0, 1), (0, -1), (2, 1), (2, -1)\).

Step 3: Calculate the second partial derivatives

To classify the critical points, calculate the second partial derivatives:\[\begin{align*} & f_{xx} = \frac{\partial^2}{\partial x^2}(x^3 + y^3 - 3x^2 - 3y + 10) = 6x - 6, \ & f_{yy} = \frac{\partial^2}{\partial y^2}(x^3 + y^3 - 3x^2 - 3y + 10) = 6y, \ & f_{xy} = \frac{\partial^2}{\partial x \partial y}(x^3 + y^3 - 3x^2 - 3y + 10) = 0.\end{align*}\]

Step 4: Apply the second derivative test

Use the second derivative test to determine the nature of each critical point. The test uses the determinant of the Hessian matrix \(D(x, y)\):\[ D = f_{xx} f_{yy} - (f_{xy})^2 \]1. For \((0, 1)\):\[\begin{align*} f_{xx} &= 6(0) - 6 = -6, \ f_{yy} &= 6(1) = 6, \ D(0, 1) &= (-6)(6) - (0)^2 = -36 \\end{align*}\]Since \(D < 0\), \((0, 1)\) is a saddle point.2. For \((0, -1)\):\[\begin{align*} f_{xx} &= -6, \ f_{yy} &= 6(-1) = -6, \ D(0, -1) &= (-6)(-6) - (0)^2 = 36 \\end{align*}\]Since \(D > 0\) and \(f_{xx} < 0\), \((0, -1)\) is a local maximum.3. For \((2, 1)\):\[\begin{align*} f_{xx} &= 6(2) - 6 = 6, \ f_{yy} &= 6(1) = 6, \ D(2, 1) &= (6)(6) - (0)^2 = 36 \\end{align*}\]Since \(D > 0\) and \(f_{xx} > 0\), \((2, 1)\) is a local minimum.4. For \((2, -1)\):\[\begin{align*} f_{xx} &= 6, \ f_{yy} &= -6, \ D(2, -1) &= (6)(-6) - (0)^2 = -36 \\end{align*}\]Since \(D < 0\), \((2, -1)\) is a saddle point.

Key Concepts

Critical Points

In multivariable calculus, finding critical points involves examining a function to determine where its rate of change is zero. These points are where a function might have a local minimum, local maximum, or a saddle point. For the function given, \(f(x, y) = x^3 + y^3 - 3x^2 - 3y + 10\), critical points are calculated by setting partial derivatives to zero.

A critical point is found where both partial derivatives of the function equal zero. This means the tangent to the surface at that point is flat in all directions on the plane. Here's the quick overview of the process:

• Compute partial derivatives of the function with respect to both variables.
• Set each of these derivatives to zero and solve the resulting system of equations.

The given exercise follows this method by setting \(f_x = 0\) and \(f_y = 0\), yielding solutions like \((0, 1), (0, -1), (2, 1), (2, -1)\) as possible critical points.

Partial Derivatives

Partial derivatives are like ordinary derivatives but focused on multivariable functions. They measure how a function changes as one variable changes while others are held constant. In this context, partial derivatives help identify critical points by showing where the gradient, the multivariable equivalent of a slope, is zero.

The function \(f(x, y)\) has two partial derivatives, one for each variable:

• The partial derivative with respect to \(x\), denoted \(f_x\), shows how \(f\) changes as only \(x\) changes.
• The partial derivative with respect to \(y\), denoted \(f_y\), shows how \(f\) changes as only \(y\) changes.

For example, in the exercise:

• \(f_x = 3x^2 - 6x\) signifies how \(f\) changes in response to a change in \(x\) with \(y\) held constant.
• \(f_y = 3y^2 - 3\) signifies how \(f\) changes in response to a change in \(y\) with \(x\) held constant.

Setting these derivatives to zero assists in finding critical points as they indicate zero change in the desired direction, suggesting potential extrema or saddle points.

Second Derivative Test

The second derivative test for multivariable functions helps classify the critical points found by the first derivatives. This is achieved by using the second partial derivatives to explore concavity and discern types of extrema. For functions of two variables, this involves calculating the Hessian matrix and its determinant.

The Hessian matrix for function \(f(x, y)\) consists of:

• \(f_{xx}\), the second derivative with respect to \(x\), showing curvature in the \(x\) direction.
• \(f_{yy}\), the second derivative with respect to \(y\), showing curvature in the \(y\) direction.
• \(f_{xy}\), the mixed second derivative.

The determinant \(D = f_{xx} f_{yy} - (f_{xy})^2\) helps determine the nature of a critical point:

• If \(D > 0\) and \(f_{xx} > 0\), the function has a local minimum.
• If \(D > 0\) and \(f_{xx} < 0\), the function has a local maximum.
• If \(D < 0\), it is a saddle point.

In the exercise, applying this test classifies points like \((0, -1)\) as a local maximum and \((2, 1)\) as a local minimum, dependent on the sign and values of \(D\) and \(f_{xx}\).