Problem 9 - 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. $$ f(x, y)=3 x^{2}-3 x y^{2}+y^{3}+3 y^{2} $$

Step-by-Step Solution

Answer

Critical points are (0, 0) - relative minimum, (2, 2) - saddle point, (1/2, -1) - saddle point.

Step 1: Find the First Partial Derivatives

To locate critical points, we first find the first partial derivatives of the function \( f(x, y) = 3x^2 - 3xy^2 + y^3 + 3y^2 \). The partial derivative with respect to \( x \) is \( f_x = \frac{\partial f}{\partial x} = 6x - 3y^2 \). The partial derivative with respect to \( y \) is \( f_y = \frac{\partial f}{\partial y} = -6xy + 3y^2 + 6y \). We'll solve \( f_x = 0 \) and \( f_y = 0 \) to find the critical points.

Step 2: Solve the System of Equations \( f_x = 0 \) and \( f_y = 0 \)

Solve the equations \( 6x - 3y^2 = 0 \) and \( -6xy + 3y^2 + 6y = 0 \) simultaneously. The first equation implies \( 6x = 3y^2 \), or \( x = \frac{1}{2}y^2 \). Substitute \( x = \frac{1}{2}y^2 \) into the second equation \( -6(\frac{1}{2}y^2)y + 3y^2 + 6y = 0 \), simplifying to \( -3y^3 + 3y^2 + 6y = 0 \). Factoring gives \( 3y(y^2 - y - 2) = 0 \). The solutions are \( y = 0, y = 2, \text{ or } y = -1 \). Substituting each value into \( x = \frac{1}{2}y^2 \), the critical points are \( (0, 0) \), \( (2, 2) \), and \( (\frac{1}{2}, -1) \).

Step 3: Find the Second Partial Derivatives

Calculate the second partial derivatives needed for the second derivative test. \( f_{xx} = \frac{\partial^2 f}{\partial x^2} = 6 \), \( f_{yy} = \frac{\partial^2 f}{\partial y^2} = 6y + 6 \), and \( f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = -6y \).

Step 4: Use the Second Derivative Test

For each critical point, use the second derivative test: Compute \( D = f_{xx}f_{yy} - (f_{xy})^2 \). - For \( (0, 0) \): \( D = (6)(6(0) + 6) - 0 = 36 \), \( f_{xx} = 6 > 0 \) so \( (0, 0) \) is a relative minimum.- For \( (2, 2) \): \( D = (6)(6(2) + 6) - (-12)^2 = 0 \), making \( (2, 2) \) a saddle point.- For \( (\frac{1}{2}, -1) \): \( D = (6)(6(-1) + 6) - 6^2 = -36 \), indicating \( (\frac{1}{2}, -1)\) is a saddle point.

Key Concepts

Partial Derivatives

Understanding partial derivatives is crucial, especially when dealing with functions of multiple variables like our given function, \( f(x, y) = 3x^2 - 3xy^2 + y^3 + 3y^2 \). Essentially, a partial derivative measures how the function changes as one of the variables changes, with all other variables held constant.
Let's break it down:

• **Partial derivative with respect to \( x \):** This gives us \( f_x = \frac{\partial f}{\partial x} = 6x - 3y^2 \). This derivative tells us how \( f(x, y) \) changes when we slightly change \( x \), keeping \( y \) constant.
• **Partial derivative with respect to \( y \):** For \( f(y) = \frac{\partial f}{\partial y} = -6xy + 3y^2 + 6y \), it shows how the function changes if we vary \( y \), with \( x \) unchanged.

Finding where both derivatives equal zero \((f_x = 0\) and \(f_y = 0)\) helps locate the critical points of the function. This is because such points mark regions in the function where the slope is zero — potential spots for maxima, minima, or saddle points.

Second Derivative Test

The second derivative test is used to classify critical points. After finding the first partial derivatives and solving for where they equal zero (our critical points), the second derivative test lets us understand the nature of these points.
To conduct this test, we calculate:

• \( f_{xx} \): The second partial derivative with respect to \( x \).
• \( f_{yy} \): The second partial derivative with respect to \( y \).
• \( f_{xy} \): The mixed partial derivative.
• \( D = f_{xx}f_{yy} - (f_{xy})^2 \): The discriminant used in the test.

For each critical point, depending on the value of \( D \):

• If \( D > 0 \) and \( f_{xx} > 0 \), it's a relative minimum.
• If \( D > 0 \) and \( f_{xx} < 0 \), it's a relative maximum.
• If \( D < 0 \), it's a saddle point, implying a change in concavity.
• If \( D = 0 \), the test is inconclusive, and the point could be a saddle point or something else.

Relative Maximum and Minimum

When determining whether a critical point is a relative maximum or minimum, one must analyze the results of the second derivative test. Remember:

• A **relative maximum** occurs where the function transitions from increasing to decreasing. In the context of the second derivative test, if \( D > 0 \) and \( f_{xx} < 0 \), the critical point is a local peak.
• A **relative minimum** is the opposite, where the function shifts from decreasing to increasing. With the test, if \( D > 0 \) and \( f_{xx} > 0 \), it's a local trough.

It's important to note that these assessments are "relative" to nearby points on the function, meaning these points are higher or lower than the immediate surroundings but not necessarily the highest or lowest overall.

Saddle Points

Saddle points are fascinating because they mark places where the function behaves interestingly. They aren't extremes (like a maximum or minimum), but rather they're points where the function curves upwards in one direction and downwards in another. Think of a horse's saddle—hence the name.
In our analysis, if \( D < 0 \), the critical point is a saddle point. This means the surface of the function zigzags through that point. Often, a graph can give insight by showing these curves, which appear to cross each way over the saddle point directionally. Thus, although these points aren't relative extrema, they are still critical for understanding the function's topography.
Be aware: at a saddle point, our function can change behavior quite sharply, which can be critical in applied fields like optimization or economics. Recognizing these points through the second derivative test can provide significant depth in understanding how a function behaves across its domain.