Critical Points: Finding & Classifying Z = (x^2-5x)(y^2-4y)

by Alex Johnson 60 views

Hey there, math enthusiasts! Today, we're diving deep into the fascinating world of multivariable calculus to explore how to find and classify critical points. We'll be tackling the function z = (x² - 5x)(y² - 4y). Buckle up, because this journey involves partial derivatives, the Hessian matrix, and a bit of algebraic finesse! Understanding critical points is crucial in various fields, from optimization problems in engineering to economic modeling, so let's get started.

1. Understanding Critical Points

Before we jump into the calculations, let's take a moment to understand what critical points actually are. In simple terms, critical points are the points where the function's slope is either zero or undefined. For a multivariable function like ours, this means we're looking for points where both partial derivatives with respect to x and y are either zero or undefined. These points are potential locations of local maxima, local minima, or saddle points. Finding critical points is the first step in understanding the behavior of a function and is the cornerstone of optimization problems.

To put it mathematically, a point (a, b) is a critical point of a function z = f(x, y) if:

  • ∂z/∂x (a, b) = 0 and ∂z/∂y (a, b) = 0
  • OR ∂z/∂x (a, b) is undefined, OR ∂z/∂y (a, b) is undefined

In our case, z = (x² - 5x)(y² - 4y), which is a polynomial function, so its partial derivatives will always be defined. Therefore, we only need to focus on finding points where the partial derivatives are equal to zero. This simplifies our task, but the algebraic manipulations can still be quite involved. The concept of critical points extends beyond two-variable functions; it is fundamental in the calculus of variations and in optimization problems in higher dimensions. Understanding this concept thoroughly will significantly enhance your ability to solve complex problems in various fields. Remember, a strong foundation in the basics is essential for tackling more advanced topics.

2. Calculating Partial Derivatives

The first step in finding critical points is to calculate the partial derivatives of our function with respect to x and y. Remember, a partial derivative is simply the derivative of the function with respect to one variable, treating all other variables as constants. This might sound a bit abstract, but it's a very powerful technique that allows us to analyze the function's behavior in different directions.

2.1 Partial Derivative with Respect to x (∂z/∂x)

To find ∂z/∂x, we treat y as a constant and differentiate the function with respect to x. Using the product rule, we get:

∂z/∂x = (2x - 5)(y² - 4y)

2.2 Partial Derivative with Respect to y (∂z/∂y)

Similarly, to find ∂z/∂y, we treat x as a constant and differentiate with respect to y:

∂z/∂y = (x² - 5x)(2y - 4)

Now we have the two partial derivatives, which are essential for finding the critical points. These derivatives represent the rate of change of the function in the x and y directions, respectively. Setting these derivatives to zero will give us the points where the function has a horizontal tangent plane, which are the potential critical points. Calculating partial derivatives accurately is crucial because any mistake here will propagate through the rest of the solution. It's always a good idea to double-check your calculations to ensure accuracy. Practice makes perfect, so the more you work with partial derivatives, the more comfortable you'll become with them.

3. Finding Critical Points

Now that we have our partial derivatives, the next step is to find the critical points. As we discussed earlier, critical points occur where both partial derivatives are equal to zero. So, we need to solve the following system of equations:

  1. (2x - 5)(y² - 4y) = 0
  2. (x² - 5x)(2y - 4) = 0

Let's analyze each equation separately to find the possible values of x and y.

3.1 Analyzing the First Equation

The first equation, (2x - 5)(y² - 4y) = 0, implies that either (2x - 5) = 0 or (y² - 4y) = 0. Let's solve each case:

  • If 2x - 5 = 0, then x = 5/2
  • If y² - 4y = 0, then y(y - 4) = 0, which gives us y = 0 or y = 4

3.2 Analyzing the Second Equation

The second equation, (x² - 5x)(2y - 4) = 0, implies that either (x² - 5x) = 0 or (2y - 4) = 0. Let's solve each case:

  • If x² - 5x = 0, then x(x - 5) = 0, which gives us x = 0 or x = 5
  • If 2y - 4 = 0, then y = 2

3.3 Combining the Results

Now we need to combine these results to find all possible pairs (x, y) that satisfy both equations. We have the following cases:

  • x = 5/2, y = 2
  • x = 0, y = 0
  • x = 0, y = 4
  • x = 5, y = 0
  • x = 5, y = 4

Therefore, the critical points are (5/2, 2), (0, 0), (0, 4), (5, 0), and (5, 4). These points are the candidates for local maxima, local minima, or saddle points. We've successfully navigated the algebraic maze to find these crucial points. The next step is to classify them, which will tell us the nature of the function's behavior around these points. This is where the Hessian matrix comes into play.

4. Classifying Critical Points Using the Hessian Matrix

Now that we've found the critical points, the next crucial step is to classify them. This means determining whether each point corresponds to a local maximum, a local minimum, or a saddle point. The Hessian matrix is our tool of choice for this task. The Hessian matrix is a matrix of second-order partial derivatives, and its properties at a critical point can tell us a lot about the function's behavior in the vicinity of that point.

4.1 Constructing the Hessian Matrix

The Hessian matrix, denoted by H, is defined as:

H = | ∂²z/∂x² ∂²z/∂x∂y | | ∂²z/∂y∂x ∂²z/∂y² |

We need to calculate the second-order partial derivatives:

  • ∂²z/∂x² = ∂/∂x (∂z/∂x) = ∂/∂x [(2x - 5)(y² - 4y)] = 2(y² - 4y)
  • ∂²z/∂y² = ∂/∂y (∂z/∂y) = ∂/∂y [(x² - 5x)(2y - 4)] = 2(x² - 5x)
  • ∂²z/∂x∂y = ∂/∂y (∂z/∂x) = ∂/∂y [(2x - 5)(y² - 4y)] = (2x - 5)(2y - 4)
  • ∂²z/∂y∂x = ∂/∂x (∂z/∂y) = ∂/∂x [(x² - 5x)(2y - 4)] = (2x - 5)(2y - 4)

Notice that ∂²z/∂x∂y = ∂²z/∂y∂x, which is expected due to Clairaut's theorem (provided the second partial derivatives are continuous, which is true for our function). Now we can write the Hessian matrix:

H = | 2(y² - 4y) (2x - 5)(2y - 4) | | (2x - 5)(2y - 4) 2(x² - 5x) |

4.2 The Discriminant

The discriminant, D, is the determinant of the Hessian matrix:

D = (∂²z/∂x²)(∂²z/∂y²) - (∂²z/∂x∂y)²

D = [2(y² - 4y)][2(x² - 5x)] - [(2x - 5)(2y - 4)]²

4.3 Classification Criteria

Now we use the following criteria to classify the critical points:

  • If D > 0 and ∂²z/∂x² > 0, then the point is a local minimum.
  • If D > 0 and ∂²z/∂x² < 0, then the point is a local maximum.
  • If D < 0, then the point is a saddle point.
  • If D = 0, the test is inconclusive, and we need to use other methods to classify the point.

5. Classifying Each Critical Point

Now we will evaluate the discriminant D and ∂²z/∂x² at each critical point to classify them.

5.1 Critical Point (0, 0)

  • ∂²z/∂x²(0, 0) = 2(0² - 4 * 0) = 0
  • ∂²z/∂y²(0, 0) = 2(0² - 5 * 0) = 0
  • ∂²z/∂x∂y(0, 0) = (2 * 0 - 5)(2 * 0 - 4) = 20
  • D(0, 0) = (0)(0) - (20)² = -400

Since D < 0, the point (0, 0) is a saddle point.

5.2 Critical Point (0, 4)

  • ∂²z/∂x²(0, 4) = 2(4² - 4 * 4) = 0
  • ∂²z/∂y²(0, 4) = 2(0² - 5 * 0) = 0
  • ∂²z/∂x∂y(0, 4) = (2 * 0 - 5)(2 * 4 - 4) = -20
  • D(0, 4) = (0)(0) - (-20)² = -400

Since D < 0, the point (0, 4) is a saddle point.

5.3 Critical Point (5, 0)

  • ∂²z/∂x²(5, 0) = 2(0² - 4 * 0) = 0
  • ∂²z/∂y²(5, 0) = 2(5² - 5 * 5) = 0
  • ∂²z/∂x∂y(5, 0) = (2 * 5 - 5)(2 * 0 - 4) = -20
  • D(5, 0) = (0)(0) - (-20)² = -400

Since D < 0, the point (5, 0) is a saddle point.

5.4 Critical Point (5, 4)

  • ∂²z/∂x²(5, 4) = 2(4² - 4 * 4) = 0
  • ∂²z/∂y²(5, 4) = 2(5² - 5 * 5) = 0
  • ∂²z/∂x∂y(5, 4) = (2 * 5 - 5)(2 * 4 - 4) = 20
  • D(5, 4) = (0)(0) - (20)² = -400

Since D < 0, the point (5, 4) is a saddle point.

5.5 Critical Point (5/2, 2)

  • ∂²z/∂x²(5/2, 2) = 2(2² - 4 * 2) = -8
  • ∂²z/∂y²(5/2, 2) = 2((5/2)² - 5 * (5/2)) = -25/2
  • ∂²z/∂x∂y(5/2, 2) = (2 * (5/2) - 5)(2 * 2 - 4) = 0
  • D(5/2, 2) = (-8)(-25/2) - (0)² = 100

Since D > 0 and ∂²z/∂x² < 0, the point (5/2, 2) is a local maximum.

6. Conclusion

In this article, we've successfully navigated the process of finding and classifying the critical points of the function z = (x² - 5x)(y² - 4y). We found the critical points to be (5/2, 2), (0, 0), (0, 4), (5, 0), and (5, 4). Using the Hessian matrix and the discriminant, we classified these points as follows:

  • (0, 0): Saddle point
  • (0, 4): Saddle point
  • (5, 0): Saddle point
  • (5, 4): Saddle point
  • (5/2, 2): Local maximum

Understanding how to find and classify critical points is essential for anyone working with multivariable functions. It's a fundamental concept in calculus with wide-ranging applications in various fields. Keep practicing, and you'll master this skill in no time! If you are interested in learning more about multivariable calculus, you can visit Khan Academy's Multivariable Calculus for more resources and practice problems.