Chapter 6: Chapter 6

Problem 152

Find the critical points and the nature of each critical point (i.e., relative maximum, relative minimum, or saddle point) for: a) $f(x, y)=x^{2}-2 x y+2 y^{2}+x-5$ b) $\mathrm{f}(\mathrm{x}, \mathrm{y})=(1-\mathrm{x})(1-\mathrm{y})(\mathrm{x}+\mathrm{y}-1)$.

Short Answer

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The critical points and their nature for the given functions are: a) For $f(x, y)=x^{2}-2 x y+2 y^{2}+x-5$, there is one critical point at $\left(-\frac{1}{2}, 0\right)$, which is a relative minimum. b) For $f(x, y)=(1-x)(1-y)(x+y-1)$, there are two critical points at $(0,0)$ and $(1,1)$. However, the second partial derivatives test is inconclusive, and we cannot determine their nature using this method.

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TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Find first-order partial derivatives

Compute the partial derivatives of the function with respect to x and y. \[f_x = \frac{\partial f}{\partial x} = 2x - 2y + 1\] \[f_y = \frac{\partial f}{\partial y} = -2x + 4y - 1\]

Step 2: Set partial derivatives equal to zero

To find the critical points, we need to solve the following system of equations: \[2x - 2y + 1= 0\] \[-2x + 4y - 1= 0\]

Step 3: Solve the system of equations

We can solve the system of equations by substitution or elimination method. Here, we use the elimination method. \[2(2x - 2y + 1) + (-2x + 4y - 1) = 0\] \[4x - 4y + 2 - 2x + 4y - 1 = 0\] \[2x + 1 = 0\] \[x = -\frac{1}{2}\] Now, substitute x back into one of the original equations: \[-2\left(-\frac{1}{2}\right) + 4y - 1= 0\] \[1 + 4y - 1= 0\] \[4y = 0\] \[y = 0\] The critical point is $\left(-\frac{1}{2}, 0\right)$.

Step 4: Analyze the critical point

To determine the nature of the critical point, we will compute the second-order partial derivatives and use the discriminant test. Second-order partial derivatives: \[f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2\] \[f_{xy} = \frac{\partial^2 f}{\partial x\partial y} = -2\] \[f_{yy} = \frac{\partial^2 f}{\partial y^2} = 4\] Discriminant: \[D = f_{xx}f_{yy} - f_{xy}^2 = (2)(4) - (-2)^2 = 8 - 4 = 4\] Since $D > 0$ and $f_{xx} > 0$, the critical point $\left(-\frac{1}{2}, 0\right)$ is a relative minimum. b) $f(x, y)=(1-x)(1-y)(x+y-1)$

Step 1: Find first-order partial derivatives

Compute the partial derivatives of the function with respect to x and y. \[f_x = (1 - y)(x + y - 1) - (1 - x)(1 - y)\] \[f_y = (1 - x)(x + y - 1) - (1 - x)(1 - y)\]

Step 2: Set partial derivatives equal to zero

To find the critical points, we need to solve the following system of equations: \[(1 - y)(x + y - 1) - (1 - x)(1 - y)= 0\] \[(1 - x)(x + y - 1) - (1 - x)(1 - y)= 0\]

Step 3: Solve the system of equations

We can solve the system of equations by expressing x and y in terms of a single variable. Since both equations look almost the same, we can use the similarity to equate them and solve for the unknown variable. Here, we equate both equations: \[(1 - y)(x + y - 1) - (1 - x)(1 - y)= (1 - x)(x + y - 1) - (1 - x)(1 - y)\] Simplifying the equation, we get: \[x = y\] Now, substitute x = y back into one of the original equations: \[ (1 - y)(y + y - 1) - (1 - y)(1 - y) = 0\] \[2y(1 - y) - (1 - y)^2 = 0\] Solve for y: \[y(2 - 2y - 1 + 2y) = (1 - y)(1 - y)= 0\] \[y = 0 \quad \text{or} \quad y = 1\] The critical points are $(0,0)$ and $(1,1)$.

Step 4: Analyze the critical points

To determine the nature of the critical points, we will compute the second-order partial derivatives and use the discriminant test. First, simplify the expressions for the first-order derivatives: \[f_x = x - y\] \[f_y = y - x\] Second-order partial derivatives: \[f_{xx} = \frac{\partial^2 f}{\partial x^2} = 1\] \[f_{xy} = \frac{\partial^2 f}{\partial x\partial y} = -1\] \[f_{yy} = \frac{\partial^2 f}{\partial y^2} = 1\] Discriminant: \[D = f_{xx}f_{yy} - f_{xy}^2 = (1)(1) - (-1)^2 = 1 - 1 = 0\] Since $D = 0$, the discriminant test is inconclusive. We cannot determine the nature of the critical points $(0,0)$ and $(1,1)$ using the second partial derivatives.

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Chapter 6: Chapter 6

Find the critical points and the nature of each critical point (i.e., relative maximum, relative minimum, or saddle point) for: a) \(f(x, y)=x^{2}-2 x y+2 y^{2}+x-5\) b) $\mathrm{f}(\mathrm{x}, \mathrm{y})=(1-\mathrm{x})(1-\mathrm{y})(\mathrm{x}+\mathrm{y}-1)$.

Short Answer

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Step 1: Find first-order partial derivatives

Step 2: Set partial derivatives equal to zero

Step 3: Solve the system of equations

Step 4: Analyze the critical point

Step 1: Find first-order partial derivatives

Step 2: Set partial derivatives equal to zero

Step 3: Solve the system of equations

Step 4: Analyze the critical points

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Chapter 1

Chapter 2

Chapter 3

Chapter 4

Chapter 5

Chapter 7

Chapter 8

Chapter 9

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 14

Chapter 15

Chapter 16

Chapter 17

Chapter 18

Chapter 19

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