Finding Critical Points Of F(x, Y) = 10x³ - 9xy² - 269x + 18y + 15 A Step-by-Step Guide

Jul 10, 2025 by ADMIN 88 views

Find All Critical Points of f(x, y) = 10x³ - 9xy² - 269x + 18y + 15

Finding critical points of a multivariable function is a fundamental concept in calculus, particularly in optimization problems. Critical points are the points where the function's first partial derivatives are simultaneously zero or undefined. These points are crucial because they represent potential local maxima, local minima, or saddle points of the function. This article will guide you through the process of finding the critical points of the function f(x, y) = 10x³ - 9xy² - 269x + 18y + 15. We'll cover the necessary steps, including calculating partial derivatives, setting up a system of equations, and solving for the critical points.

Understanding Critical Points

Before diving into the calculations, let's clarify what critical points are and why they are important. In single-variable calculus, a critical point of a function f(x) is a point c in the domain of f where either f'(c) = 0 or f'(c) is undefined. In multivariable calculus, the concept is similar, but we consider partial derivatives.

For a function f(x, y), the critical points are the solutions to the system of equations:

∂f/∂x = 0

∂f/∂y = 0

These equations represent the points where the tangent plane to the surface z = f(x, y) is horizontal. These points are potential locations of local extrema (maxima or minima) or saddle points.

Step-by-Step Solution

To find the critical points of f(x, y) = 10x³ - 9xy² - 269x + 18y + 15, we'll follow these steps:

Calculate the partial derivative of f with respect to x, denoted as ∂f/∂x.
Calculate the partial derivative of f with respect to y, denoted as ∂f/∂y.
Set both partial derivatives equal to zero.
Solve the resulting system of equations for x and y.
The solutions (x, y) are the critical points of the function.

1. Calculate the Partial Derivative with Respect to x (∂f/∂x)

To find ∂f/∂x, we differentiate f(x, y) with respect to x, treating y as a constant:

∂f/∂x = ∂/∂x (10x³ - 9xy² - 269x + 18y + 15)

Applying the power rule and the constant multiple rule, we get:

∂f/∂x = 30x² - 9y² - 269

2. Calculate the Partial Derivative with Respect to y (∂f/∂y)

Next, we find ∂f/∂y by differentiating f(x, y) with respect to y, treating x as a constant:

∂f/∂y = ∂/∂y (10x³ - 9xy² - 269x + 18y + 15)

Applying the power rule and the constant multiple rule, we get:

∂f/∂y = -18xy + 18

3. Set Both Partial Derivatives Equal to Zero

Now, we set both partial derivatives equal to zero to find the points where the tangent plane is horizontal:

30x² - 9y² - 269 = 0 (Equation 1)

-18xy + 18 = 0 (Equation 2)

4. Solve the System of Equations

We now have a system of two equations with two variables, x and y. Let's solve this system.

From Equation 2, we have:

-18xy + 18 = 0

18xy = 18

xy = 1

If x=0 or y=0 the equation is not valid. Thus, the valid solutions should have x≠0 and y≠0. We can express y in terms of x:

y = 1/x

Now, substitute this expression for y into Equation 1:

30x² - 9(1/x)² - 269 = 0

30x² - 9/x² - 269 = 0

Multiply through by x² to eliminate the fraction:

30x⁴ - 9 - 269x² = 0

Rearrange the equation:

30x⁴ - 269x² - 9 = 0

This is a quadratic equation in x². Let z = x², then the equation becomes:

30z² - 269z - 9 = 0

We can solve this quadratic equation for z using the quadratic formula:

z = (-b ± √(b² - 4ac)) / (2a)

Where a = 30, b = -269, and c = -9.

z = (269 ± √((-269)² - 4(30)(-9))) / (2(30))

z = (269 ± √(72361 + 1080)) / 60

z = (269 ± √73441) / 60

z = (269 ± 271) / 60

We have two possible values for z:

z₁ = (269 + 271) / 60 = 540 / 60 = 9

z₂ = (269 - 271) / 60 = -2 / 60 = -1/30

Since z = x², and x is a real number, z must be non-negative. Therefore, z₂ = -1/30 is not a valid solution. Thus, we have:

x² = 9

x = ±3

Now, we find the corresponding values of y using y = 1/x:

If x = 3, then y = 1/3.

If x = -3, then y = -1/3.

5. Identify the Critical Points

The critical points are the pairs (x, y) that satisfy the system of equations. We have found two critical points:

(3, 1/3)

(-3, -1/3)

Conclusion

We have successfully found the critical points of the function f(x, y) = 10x³ - 9xy² - 269x + 18y + 15. The critical points are (3, 1/3) and (-3, -1/3). These points are crucial for further analysis, such as determining whether they correspond to local maxima, local minima, or saddle points using the second derivative test or other methods. Understanding how to find critical points is essential in various applications, including optimization problems in physics, engineering, and economics.

By following the steps outlined in this article, you can confidently find the critical points of similar multivariable functions. Remember to carefully calculate the partial derivatives, set up the system of equations, and solve for the variables. This process is a cornerstone of multivariable calculus and is vital for understanding the behavior of functions in higher dimensions.

In the realm of multivariable calculus, the identification of critical points stands as a cornerstone concept, pivotal for unraveling the behavior of functions and their graphical representations. Critical points serve as the potential locales for local maxima, local minima, or saddle points on a function's surface. In this exposition, we embark on a comprehensive exploration of critical points, delving into their definition, the methodology for their determination, and their significance in mathematical analysis and optimization problems.

Defining Critical Points

At its core, a critical point of a function f(x, y) represents a juncture in the domain where the first partial derivatives of the function simultaneously vanish or remain undefined. Mathematically, this condition is expressed as:

∂f/∂x = 0

∂f/∂y = 0

Here, ∂f/∂x and ∂f/∂y denote the partial derivatives of f with respect to x and y, respectively. The satisfaction of these equations implies that the tangent plane to the surface z = f(x, y) is horizontal at the point under consideration. This horizontal tangency hints at the possibility of an extremum (maximum or minimum) or a saddle point lurking in the vicinity.

Methodology for Determining Critical Points

The pursuit of critical points entails a systematic approach, encompassing the following steps:

Compute Partial Derivatives: The initial stride involves the computation of the first partial derivatives of f(x, y) with respect to both x and y. These derivatives encapsulate the rate of change of f along the x and y axes, respectively.
Equate to Zero: Subsequently, we set each of the partial derivatives equal to zero, thereby yielding a system of equations. The solutions to this system are poised to unveil the critical points of the function.
Solve the System: The ensuing challenge lies in solving the system of equations obtained in the preceding step. This endeavor may necessitate a repertoire of algebraic techniques, encompassing substitution, elimination, or numerical methods, depending on the complexity of the equations.
Identify Critical Points: Once the solutions (x, y) to the system of equations are secured, these points epitomize the critical points of the function. Each critical point encapsulates a potential locale for extrema or saddle points.

Significance of Critical Points

Critical points wield considerable significance in mathematical analysis and optimization problems, serving as pivotal landmarks in the landscape of functions. Their importance stems from the following facets:

Extrema Identification: Critical points serve as the primary candidates for extrema (maxima and minima) of the function. By scrutinizing the function's behavior in the vicinity of critical points, we can discern whether they correspond to local maxima, local minima, or neither. This analysis often involves the application of the second derivative test or other techniques tailored for extremum classification.
Optimization Problems: In the realm of optimization, where the goal is to identify the optimal values of a function subject to constraints, critical points assume paramount importance. They pinpoint the potential solutions that warrant further investigation. By focusing on critical points, we can streamline the search for optimal solutions, enhancing efficiency and accuracy.
Graphical Analysis: Critical points enrich our comprehension of a function's graph by demarcating key features such as peaks, valleys, and saddle points. These features offer valuable insights into the function's overall shape and behavior, facilitating visualization and interpretation. The identification of critical points thus augments our ability to sketch and analyze complex functions.

Illustrative Example

To solidify the understanding of critical points, let's consider an illustrative example. Suppose we seek the critical points of the function:

f(x, y) = x³ - 3xy + y³

Compute Partial Derivatives: We initiate by computing the partial derivatives of f with respect to x and y:

∂f/∂x = 3x² - 3y

∂f/∂y = -3x + 3y²

Equate to Zero: Next, we set these partial derivatives equal to zero:

3x² - 3y = 0

-3x + 3y² = 0

Solve the System: To solve this system, we can employ substitution or elimination methods. From the first equation, we have:

y = x²

Substituting this into the second equation, we get:

-3x + 3(x²)² = 0

-3x + 3x⁴ = 0

x(x³ - 1) = 0

This equation yields two solutions for x: x = 0 and x = 1. The corresponding values for y are obtained by substituting these x values back into the equation y = x²:

For x = 0, we have y = 0² = 0.

For x = 1, we have y = 1² = 1.

Identify Critical Points: Hence, the critical points of the function f(x, y) = x³ - 3xy + y³ are (0, 0) and (1, 1).

Conclusion

Critical points constitute a pivotal concept in multivariable calculus, serving as potential hubs for extrema and saddle points. Their determination involves the computation of partial derivatives, the formulation of a system of equations, and the solution of this system. These points play a crucial role in optimization problems, graphical analysis, and the broader understanding of function behavior. By mastering the art of identifying critical points, mathematicians and practitioners alike can gain deeper insights into the intricacies of multivariable functions and their applications across diverse domains.

Critical points in multivariable calculus are fundamental concepts that help us understand the behavior of functions, especially in the context of optimization and graphical analysis. They represent locations where the function's rate of change is zero or undefined, making them potential candidates for local maxima, local minima, or saddle points. This detailed guide provides an in-depth exploration of how to find critical points for a given multivariable function, highlighting the steps, techniques, and significance of this process. Understanding this concept is crucial for various applications in mathematics, physics, engineering, and economics, where optimization is a key goal.

Understanding the Basics of Critical Points

In single-variable calculus, a critical point of a function f(x) is a point c in the domain of f where the derivative f'(c) is either zero or undefined. This concept extends to multivariable calculus, but instead of a single derivative, we deal with partial derivatives.

For a function f(x, y) of two variables, a critical point (a, b) is a point where both the first partial derivatives are either zero or do not exist. Mathematically, this means:

∂f/∂x (a, b) = 0 or undefined

∂f/∂y (a, b) = 0 or undefined

The partial derivatives represent the rate of change of the function along the x and y axes, respectively. When both partial derivatives are zero, it suggests that the tangent plane to the surface z = f(x, y) is horizontal, which is a characteristic of extrema (maxima or minima) and saddle points.

Why Are Critical Points Important?

Critical points are vital for several reasons:

Optimization: In optimization problems, we often seek to find the maximum or minimum values of a function. Critical points provide a set of candidates for these extrema. By analyzing the function's behavior around these points, we can determine the local maxima, local minima, and saddle points.
Graphical Analysis: The critical points help us understand the shape and behavior of a function's graph. They identify peaks, valleys, and saddle points, which are crucial features for sketching and interpreting the function.
Applications in Science and Engineering: Many real-world problems, such as finding the optimal design parameters or minimizing costs, involve optimizing functions. Critical points are essential tools in these applications.

Step-by-Step Guide to Finding Critical Points

To find the critical points of a multivariable function f(x, y), follow these steps:

Step 1: Compute the Partial Derivatives

The first step is to compute the first partial derivatives of f(x, y) with respect to x and y. This involves differentiating the function with respect to each variable while treating the other variable as a constant.

Partial Derivative with Respect to x (∂f/∂x)

To find ∂f/∂x, differentiate f(x, y) with respect to x, treating y as a constant. For example, if f(x, y) = x³ + 3x²y - 6xy + y², then:

∂f/∂x = ∂/∂x (x³ + 3x²y - 6xy + y²)

∂f/∂x = 3x² + 6xy - 6y

Partial Derivative with Respect to y (∂f/∂y)

Similarly, to find ∂f/∂y, differentiate f(x, y) with respect to y, treating x as a constant. Using the same example function:

∂f/∂y = ∂/∂y (x³ + 3x²y - 6xy + y²)

∂f/∂y = 3x² - 6x + 2y

Step 2: Set the Partial Derivatives Equal to Zero

Next, set both partial derivatives equal to zero. This creates a system of equations that we need to solve for x and y:

∂f/∂x = 0

∂f/∂y = 0

Using our example, we have:

3x² + 6xy - 6y = 0 (Equation 1)

3x² - 6x + 2y = 0 (Equation 2)

Step 3: Solve the System of Equations

Now, solve the system of equations to find the values of x and y that satisfy both equations simultaneously. This can be done using various methods, such as substitution, elimination, or numerical techniques.

Solving by Substitution

From Equation 2, we can express y in terms of x:

2y = 6x - 3x²

y = 3x - (3/2)x²

Substitute this expression for y into Equation 1:

3x² + 6x(3x - (3/2)x²) - 6(3x - (3/2)x²) = 0

3x² + 18x² - 9x³ - 18x + 9x² = 0

-9x³ + 30x² - 18x = 0

Divide by -3x (assuming x ≠ 0):

3x² - 10x + 6 = 0

Now, solve this quadratic equation for x using the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

x = (10 ± √(100 - 4(3)(6))) / (2(3))

x = (10 ± √28) / 6

x = (10 ± 2√7) / 6

x = (5 ± √7) / 3

So, we have two values for x:

x₁ = (5 + √7) / 3

x₂ = (5 - √7) / 3

Find Corresponding y Values

Substitute these x values back into the equation y = 3x - (3/2)x² to find the corresponding y values:

For x₁ = (5 + √7) / 3:

y₁ = 3((5 + √7) / 3) - (3/2)(((5 + √7) / 3)²)

y₁ = (5 + √7) - (1/6)(25 + 10√7 + 7)

y₁ = (5 + √7) - (1/6)(32 + 10√7)

y₁ = (5 + √7) - (16/3 + (5/3)√7)

y₁ = (15 + 3√7 - 32 - 5√7) / 3

y₁ = (-17 - 2√7) / 3

For x₂ = (5 - √7) / 3:

y₂ = 3((5 - √7) / 3) - (3/2)(((5 - √7) / 3)²)

y₂ = (5 - √7) - (1/6)(25 - 10√7 + 7)

y₂ = (5 - √7) - (1/6)(32 - 10√7)

y₂ = (5 - √7) - (16/3 - (5/3)√7)

y₂ = (15 - 3√7 - 32 + 5√7) / 3

y₂ = (-17 + 2√7) / 3

Step 4: Identify the Critical Points

Based on the solutions obtained, the critical points are the pairs (x, y) that satisfy the system of equations. In this case, we have two critical points:

((5 + √7) / 3, (-17 - 2√7) / 3)

((5 - √7) / 3, (-17 + 2√7) / 3)

Special Case: x = 0

We initially divided by x, assuming x ≠ 0. Now, we need to check what happens when x = 0. If x = 0, from Equation 2:

3(0)² - 6(0) + 2y = 0

2y = 0

y = 0

So, (0, 0) is also a critical point.

Complete Set of Critical Points

In summary, the critical points for the function f(x, y) = x³ + 3x²y - 6xy + y² are:

(0, 0)

((5 + √7) / 3, (-17 - 2√7) / 3)

((5 - √7) / 3, (-17 + 2√7) / 3)

Advanced Techniques and Considerations

Second Derivative Test

Once you have identified the critical points, you can use the second derivative test to determine whether each point is a local maximum, a local minimum, or a saddle point. The second derivative test involves computing the second partial derivatives and the determinant of the Hessian matrix.

Functions with Undefined Derivatives

In some cases, the partial derivatives may not exist at certain points. These points should also be considered as critical points and analyzed separately.

Numerical Methods

For complex functions, solving the system of equations analytically may be challenging. In such cases, numerical methods, such as Newton's method or gradient descent, can be used to approximate the critical points.

Conclusion

Finding critical points of multivariable functions is a crucial skill in calculus and has wide-ranging applications. By following the steps outlined in this guide—computing partial derivatives, setting them equal to zero, solving the system of equations, and considering special cases—you can effectively identify these points. Remember that critical points are essential for understanding the behavior of functions and solving optimization problems. The ability to find and analyze these points provides valuable insights into the mathematical models used across various scientific and engineering disciplines.