Problem 8 - 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. $$ f(x, y)=x^{3}-6 x^{2}-3 y^{2} $$

Step-by-Step Solution

Answer

Critical points are \((0,0)\) (relative maximum) and \((4,0)\) (saddle point).

Step 1: Find the First Partial Derivatives

First, find the partial derivative of the function with respect to \(x\), which is \(f_x(x, y) = \frac{\partial}{\partial x}(x^3 - 6x^2 - 3y^2) = 3x^2 - 12x\). Then find the partial derivative with respect to \(y\), which is \(f_y(x, y) = \frac{\partial}{\partial y}(x^3 - 6x^2 - 3y^2) = -6y\).

Step 2: Solve for Critical Points

Set the first partial derivatives equal to zero to find critical points:\( f_x(x, y) = 3x^2 - 12x = 0 \) and \( f_y(x, y) = -6y = 0 \). The solutions to these equations are: \(x(3x - 12) = 0\) leading to \(x = 0\) or \(x = 4\) and \(y = 0\). Hence, the critical points are \((0,0)\) and \((4,0)\).

Step 3: Find the Second Partial Derivatives

Calculate the second partial derivatives: \(f_{xx}(x, y) = \frac{\partial^2}{\partial x^2}(x^3 - 6x^2 - 3y^2) = 6x - 12\), \(f_{yy}(x, y) = \frac{\partial^2}{\partial y^2}(x^3 - 6x^2 - 3y^2) = -6\), and \(f_{xy}(x, y) = \frac{\partial^2}{\partial x \partial y}(x^3 - 6x^2 - 3y^2) = 0\).

Step 4: Use Second Derivative Test

The second derivative test is applied using the determinant of the Hessian matrix: \(D = f_{xx} f_{yy} - (f_{xy})^2\). Calculate \(D\) for each critical point. For \((0,0)\): \( f_{xx}(0,0) = -12, f_{yy}(0,0) = -6, f_{xy}(0,0) = 0\). Thus, \(D = (-12)(-6) - 0^2 = 72\), which is positive, and since \(f_{xx} < 0\), \((0,0)\) is a relative maximum.For \((4,0)\): \( f_{xx}(4,0) = 12, f_{yy}(4,0) = -6, f_{xy}(4,0) = 0\). Thus, \(D = (12)(-6) - 0^2 = -72\), which is negative, indicating \((4,0)\) is a saddle point.

Key Concepts

Partial Derivatives

When dealing with functions of several variables, like our function \( f(x, y) = x^3 - 6x^2 - 3y^2 \), we often want to understand how the function changes along the different axes. This is where partial derivatives come into play. They measure how the function changes as we make small changes only in one variable, while keeping the others constant.

• The partial derivative with respect to \( x \) is represented as \( f_x \), and it shows how \( f \) changes as \( x \) changes, while \( y \) is held constant. For our function, it is \( f_x(x, y) = 3x^2 - 12x \).
• The partial derivative with respect to \( y \) is represented as \( f_y \), and it shows how \( f \) changes as \( y \) changes, while \( x \) is held constant. For our function, it is \( f_y(x, y) = -6y \).

To find critical points, we set these first partial derivatives to zero because it helps us find where the function does not increase or decrease, indicating potential maxima, minima, or saddle points.

Relative Maximum

A relative maximum is a point where a function has a peak compared to points nearby. In simpler terms, in a small neighborhood around this point, the function's value is higher than at nearby points. To determine if a point is a relative maximum, we use the second derivative test.
For our function \( f(x, y) = x^3 - 6x^2 - 3y^2 \), the critical point \( (0, 0) \) was found using the first partial derivatives.
Applying the second derivative test, we calculate:

• \( f_{xx} = 6x - 12 \) at \( (0,0) \) gives us \( -12 \).
• \( f_{yy} = -6 \) remains \(-6\) regardless of \( x \) and \( y \).
• The determinant \( D = f_{xx} f_{yy} - (f_{xy})^2 \), and for \((0,0)\), \( D = 72 \).

Since \( D > 0 \) and \( f_{xx} < 0 \), this confirms that \( (0, 0) \) is a relative maximum.

Relative Minimum

A relative minimum is the opposite of a relative maximum. It's a point where the function dips compared to nearby points. In a neighborhood around this point, the function's value is lower than at places nearby.
To determine if a point is a relative minimum, we apply the second derivative test. However, for this particular function, after performing the test on all critical points, no relative minima were found.
Here's how it generally works: if \( D > 0 \) (determined from the Hessian matrix) and \( f_{xx}(x, y) > 0 \), then a point is a relative minimum. In our case, neither of the points tested, \((0,0)\) or \((4,0)\), met both these conditions.

Saddle Point

Saddle points are interesting because they aren't local maxima or minima. The graph at these points looks like a "saddle," climbing in one direction and falling in another; some directions increase while others decrease.
For our function \( f(x, y) = x^3 - 6x^2 - 3y^2 \), the second derivative test reveals that the point \( (4, 0) \) is a saddle point.

• Evaluating the second derivatives at \( (4,0) \) gives \( f_{xx}(4,0) = 12 \), \( f_{yy}(4,0) = -6 \), and a cross term \( f_{xy}(4,0) = 0 \).
• The determinant \( D = (12)(-6) - 0^2 = -72 \) is less than zero.

A \( D < 0 \) indicates that \( (4, 0) \) is a saddle point, showing the wonky, neither-peaking-nor-dipping behavior characteristic of saddle points.