Explanation Given information: The function of two variable, f ( x , y ) = ( x − 1 ) ( y − 1 ) ( x + y − 1 ) . Formula used: The second order partial derivatives for a function of two variables z = g ( x , y ) are as follows: With respect to x , ∂ 2 f ∂ x 2 g ( x , y ) = ∂ ∂ x ( ∂ g ∂ x ) = ( g x ) x = g x x With respect to y , ∂ 2 f ∂ y 2 g ( x , y ) = ∂ ∂ y ( ∂ g ∂ y ) = ( g y ) y = g y y The mixed second order partial derivatives for a function of two variables z = g ( x , y ) are as follows: ∂ 2 f ∂ x ∂ y g ( x , y ) = ∂ ∂ x ( ∂ g ∂ y ) = ( g y ) x = g y x And ∂ 2 f ∂ y ∂ x g ( x , y ) = ∂ ∂ y ( ∂ g ∂ x ) = ( g x ) y = g x y The discriminant D for a function of two variable is used to classify whether the critical points act as relative maximum, relative minimum or saddle point. D = f x x f y y − ( f x y ) 2 Suppose that ( a , b ) is a critical point for the function g ( x , y ) , then we evaluate the value of discriminant D and g x x at that point. If the sign of D is positive and that of g x x is positive then the function has relative minimum at that critical point. If the sign of D is negative then the function has saddle point. If the sign of D is positive and that of g x x is negative then the function has relative maximum at that critical point. If the D = 0 then we cannot conclude anything. Calculation: Consider the function, f ( x , y ) = ( x − 1 ) ( y − 1 ) ( x + y − 1 ) . Convert the equation in the standard form, that is, f ( x , y ) = ( x − 1 ) ( y − 1 ) ( x + y − 1 ) = x 2 y + x y 2 − 3 x y − x 2 − y 2 + 2 x + 2 y − 1 Partially differentiate the function with respect to x and y to find the critical points. Partial differentiation with respect to x, ∂ ∂ x f ( x , y ) = ∂ ∂ x ( x 2 y + x y 2 − 3 x y − x 2 − y 2 + 2 x + 2 y − 1 ) = 2 x y + y 2 − 3 y − 2 x + 2 Partial differentiation with respect to y, ∂ ∂ y f ( x , y ) = ∂ ∂ y ( x 2 y + x y 2 − 3 x y − x 2 − y 2 + 2 x + 2 y − 1 ) = 2 x y + x 2 − 3 x − 2 y + 2 Equate the first order derivatives to 0 and solve for x and y, { 2 x y + y 2 − 3 y − 2 x + 2 = 0 2 x y + x 2 − 3 x − 2 y + 2 = 0 Solving the system of equation 2 x y + y 2 − 3 y − 2 x + 2 = 0 , Subtract y 2 − 3 y + 2 from both the sides, 2 x y − 2 x = − y 2 + 3 y − 2 2 x ( y − 1 ) = − y 2 + 3 y − 2 2 x ( y − 1 ) = − y 2 + 2 y + y − 2 2 x ( y − 1 ) = − y ( y − 2 ) + 1 ( y − 2 ) Further, we can simplify it as when y ≠ 1 , 2 x ( y − 1 ) = − ( y − 2 ) ( y − 1 ) x = − y − 2 2 Put the value of x in 2 x y + x 2 − 3 x − 2 y + 2 = 0 to get value of y , We get the two values, y = 0 , 2 3 . Therefore, the values of xare 1 , 2 3 . Now, to evaluate the value of xwhen y = 1 , Put y = 1 in 2 x y + y 2 − 3 y − 2 x + 2 = 0 . We get x = 1 . And when we put x = 0 we get y = 1 . The critical points are ( 0 , 1 ) , ( 1 , 0 ) , ( 1 , 1 ) and ( 2 3 , 2 3 )

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

6th Edition

ISBN: 9781524908102

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

See similar textbooks

Concept explainers

Minimization

In mathematics, traditional optimization problems are typically expressed in terms of minimization. When we talk about minimizing or maximizing a function, we refer to the maximum and minimum possible values of that function. This can be expressed in ter…

Maxima and Minima

The extreme points of a function are the maximum and the minimum points of the function. A maximum is attained when the function takes the maximum value and a minimum is attained when the function takes the minimum value.

Derivatives

A derivative means a change. Geometrically it can be represented as a line with some steepness. Imagine climbing a mountain which is very steep and 500 meters high. Is it easier to climb? Definitely not! Suppose walking on the road for 500 meters. Which …

Concavity

In calculus, concavity is a descriptor of mathematics that tells about the shape of the graph. It is the parameter that helps to estimate the maximum and minimum value of any of the functions and the concave nature using the graphical method. We use the …

Videos

Question

Chapter 11, Problem 41SP

To determine

To calculate: The critical points of the function, f(x,y)=(x−1)(y−1)(x+y−1) and classify them as relative maximum, relative minimum or saddle point.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 11, Problem 40SP

Chapter 11, Problem 42SP

Students have asked these similar questions

2. a. Find relative extrema of the following functions: 2 9x y = x2 – 25 y = x² + 9 - Please show detail calculations. 7 b. Find absolute maximum and minimum of f (x) = –(x + 1)3 + (x + 1)3 on (-5,5).

6. Find all critical points of the given function and use the second derivative test to identify each as a local maximum, a local minimum, or undetermined. f(x, y) x² + y³ + 6x − 3y

2. a. Find relative extrema of the following functions: 9x y = x² – 25 y = x² + 9 Please show detail calculations. b. Find absolute maximum and minimum of f(x) = -(x+1) +=(x+ 1)i on (-5,5).

Chapter 11 Solutions

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

Ch. 11.1 - Prob. 1PS Ch. 11.1 - Prob. 2PS Ch. 11.1 - Prob. 3PS Ch. 11.1 - Prob. 4PS Ch. 11.1 - Prob. 5PS Ch. 11.1 - Prob. 6PS Ch. 11.1 - Prob. 7PS Ch. 11.1 - Prob. 8PS Ch. 11.1 - Prob. 9PS Ch. 11.1 - Prob. 10PS

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

The critical points of the function, f ( x , y ) = ( x − 1 ) ( y − 1 ) ( x + y − 1 ) and classify them as relative maximum , relative minimum or saddle point.

Solution Summary: The author explains the second order partial derivatives for a function of two variables z=g(x,y).

Concept explainers

Videos

To calculate: The critical points of the function, f(x,y)=(x−1)(y−1)(x+y−1) and classify them as relative maximum, relative minimum or saddle point.

Want to see the full answer?

Chapter 11 Solutions