Find And Classify The Critical Points Of Z = (x^2 + 6x) (y^2 + 4y). Local Maximums: _Local Minimums:
Understanding how to find and classify critical points of a multivariable function is a fundamental concept in calculus, especially in the context of optimization problems. The function in question, Z = (x^2 + 6x)(y^2 + 4y), presents a rich case study for applying techniques such as setting derivatives to zero, analyzing second derivatives, and understanding the nature of critical points. This article offers a comprehensive guide on how to find and classify the critical points of this function, detailing each step and providing insights into the process of determining local maximums and minimums.
Introduction to Critical Points and Their Classification
In multivariable calculus, critical points are points in the domain where the gradient vector of the function vanishes—that is, where all first-order partial derivatives are zero. These points are essential because they can be locations of local maxima, local minima, or saddle points.
Key Concepts:
- Critical Point: A point (x, y) where ∇Z = 0.
- Second Derivative Test: Used to classify the nature of a critical point based on second derivatives.
Why Classify Critical Points?
Classifying critical points helps in understanding the behavior of the function around those points, which is particularly useful in optimization problems, physics, economics, and engineering.
Step 1: Expressing the Function Clearly
Before proceeding with derivatives, it's important to write the function clearly:
\[ Z = (x^2 + 6x)(y^2 + 4y) \]
This is a product of two quadratic expressions, which suggests that the critical points can be found by analyzing the derivatives with respect to x and y separately.
Step 2: Computing Partial Derivatives
To find the critical points, we need the first-order partial derivatives of Z concerning x and y.
Partial derivative with respect to x:
Using the product rule:
\[
\frac{\partial Z}{\partial x} = \frac{\partial}{\partial x} (x^2 + 6x)(y^2 + 4y)
= (2x + 6)(y^2 + 4y)
\]
Similarly, the partial derivative with respect to y:
Partial derivative with respect to y:
\[
\frac{\partial Z}{\partial y} = (x^2 + 6x) \frac{\partial}{\partial y}(y^2 + 4y) = (x^2 + 6x)(2y + 4)
\]
Step 3: Setting Partial Derivatives Equal to Zero
Critical points occur where both partial derivatives are zero simultaneously:
\[
(2x + 6)(y^2 + 4y) = 0
\]
\[
(x^2 + 6x)(2y + 4) = 0
\]
These products equal zero when either factor in each is zero, leading to four cases.
Case 1:
\[
2x + 6 = 0 \quad \Rightarrow \quad x = -3
\]
\[
y^2 + 4y = 0 \quad \Rightarrow \quad y(y + 4) = 0 \quad \Rightarrow \quad y=0 \quad \text{or} \quad y=-4
\]
Critical points here:
- \((-3, 0)\)
- \((-3, -4)\)
Case 2:
\[
x^2 + 6x = 0 \quad \Rightarrow \quad x(x + 6) = 0 \quad \Rightarrow \quad x=0 \quad \text{or} \quad x=-6
\]
\[
2y + 4 = 0 \quad \Rightarrow \quad y = -2
\]
Critical points:
- \((0, -2)\)
- \((-6, -2)\)
Step 4: List of Critical Points
Summarizing, the critical points are:
- \((-3, 0)\)
- \((-3, -4)\)
- \((0, -2)\)
- \((-6, -2)\)
Step 5: Classifying Critical Points Using the Second Derivative Test
To classify these critical points as local maxima, minima, or saddle points, we analyze the second derivatives and compute the discriminant \( D \):
\[
D = Z{xx} Z{yy} - (Z_{xy})^2
\]
Where:
- \( Z_{xx} \) = second partial derivative with respect to x,
- \( Z_{yy} \) = second partial derivative with respect to y,
- \( Z_{xy} \) = mixed second partial derivative.
Calculating Second Derivatives:
From the first derivatives:
\[
Z_x = (2x + 6)(y^2 + 4y)
\]
\[
Z_y = (x^2 + 6x)(2y + 4)
\]
Compute second derivatives:
\[
Z{xx} = \frac{\partial}{\partial x} Zx = \frac{\partial}{\partial x} [(2x + 6)(y^2 + 4y)] = 2(y^2 + 4y)
\]
\[
Z{yy} = \frac{\partial}{\partial y} Zy = \frac{\partial}{\partial y} [(x^2 + 6x)(2y + 4)] = 2(x^2 + 6x)
\]
\[
Z{xy} = \frac{\partial}{\partial y} Zx = \frac{\partial}{\partial y} [(2x + 6)(y^2 + 4y)] = (2x + 6)(2y + 4)
\]
Now, evaluate the second derivatives at each critical point.
Step 6: Evaluating and Classifying Each Critical Point
Let's evaluate \( Z{xx} \), \( Z{yy} \), and \( Z_{xy} \) at each critical point, then compute \( D \):
---
Critical Point 1: \((-3, 0)\)
- \( Z_{xx} = 2(0^2 + 4 \times 0) = 0 \)
- \( Z_{yy} = 2((-3)^2 + 6 \times -3) = 2(9 - 18) = 2(-9) = -18 \)
- \( Z_{xy} = (2 \times -3 + 6)(2 \times 0 + 4) = ( -6 + 6 )(4) = 0 \times 4 = 0 \)
Discriminant:
\[
D = (0)(-18) - (0)^2 = 0
\]
Since D=0, the second derivative test is inconclusive. Further analysis is needed, but initial indicators suggest a saddle point or a flat point.
---
Critical Point 2: \((-3, -4)\)
- \( Z_{xx} = 2((-4)^2 + 4 \times -4) = 2(16 - 16) = 2(0) = 0 \)
- \( Z_{yy} = 2(9 + 6 \times -3) = 2(9 - 18) = 2(-9) = -18 \)
- \( Z_{xy} = ( -6 + 6 )(2 \times -4 + 4) = 0 \times (-8 + 4) = 0 \)
Discriminant:
\[
D = 0 \times -18 - 0^2 = 0
\]
Again, D=0, indicating an inconclusive classification.
---
Critical Point 3: \((0, -2)\)
- \( Z_{xx} = 2((-2)^2 + 4 \times -2) = 2(4 - 8) = 2(-4) = -8 \)
- \( Z_{yy} = 2(0^2 + 6 \times 0) = 0 \)
- \( Z_{xy} = (2 \times 0 + 6)(2 \times -2 + 4) = 6 \times (-4 + 4) = 6 \times 0 = 0 \)
Discriminant:
\[
D = (-8)(0) - 0^2 = 0
\]
Inconclusive again.
---
Critical Point 4: \((-6, -2)\)
- \( Z_{xx} = 2((-2)^2 + 4 \times -2) = 2(4 - 8) = 2(-4) = -8 \)
- \( Z_{yy} = 2(( -6)^2 + 6 \times -6) = 2(36 - 36) = 2(0) = 0 \)
- \( Z_{xy}
