Explanation Given: The function is f ( x , y ) = x 2 − 4 y 2 + x y on the region 4 x 2 + 9 y 2 ≤ 36 . Definition used: “The Lagrange multipliers defined as ∇ f ( x , y , z ) = λ ∇ g ( x , y , z ) . This equation can be expressed as f x = λ g x , f y = λ g y , f z = λ g z and g ( x , y , z ) = k ”. Calculation: Find the partial derivatives of the function f ( x , y ) = x 2 − 4 y 2 + x y . f ( x , y ) = x 2 − 4 y 2 + x y f x ( x , y ) = 2 x + y f y ( x , y ) = − 8 y + x Set the partial derivatives f x ( x , y ) = 2 x + y , f y ( x , y ) = − 8 y + x to zero and find the values of x and y . Subtract the equations (1) – 2(2), ( 2 x + y ) − 2 ( − 8 y + x ) = 0 2 x + y + 16 y − 2 x = 0 17 y = 0 y = 0 Substitute y = 0 in 2 x + y = 0 , 2 x + y = 0 2 x + 0 = 0 2 x = 0 x = 0 Thus, the critical point of the function f ( x , y ) = x 2 − 4 y 2 + x y is ( 0 , 0 ) . Substitute ( 0 , 0 ) in f ( x , y ) = x 2 − 4 y 2 + x y and obtain, f ( 0 , 0 ) = ( 0 ) 2 − 4 ( 0 ) 2 + ( 0 ) ( 0 ) = 0 − 0 + 0 = 0 Thus, the value of f ( 0 , 0 ) = 0 . This critical point ( 0 , 0 ) is inside the given region 4 x 2 + 9 y 2 ≤ 36 as it satisfies the inequality. Let the boundary function of the inequality 4 x 2 + 9 y 2 ≤ 36 be g ( x , y ) = 4 x 2 + 9 y 2 = 36 . The Lagrange multipliers ∇ f ( x , y ) = λ ∇ g ( x , y ) is computed as follows. ∇ f ( x , y ) = λ ∇ g ( x , y ) 〈 f x , f y 〉 = λ 〈 g x , g y 〉 〈 f x ( x 2 − 4 y 2 + x y ) , f y ( x 2 − 4 y 2 + x y ) 〉 = λ 〈 g x ( 4 x 2 + 9 y 2 ) , g y ( 4 x 2 + 9 y 2 ) 〉 〈 2 x + y , − 8 y + x 〉 = λ 〈 8 x , 18 y 〉 The result 〈 2 x + y , − 8 y + x 〉 = λ 〈 8 x , 18 y 〉 , can be expressed as follows. 2 x + y = λ ( 8 x ) (1) − 8 y + x = λ ( 18 y ) (2) From the equation (1), 2 x + y = λ ( 8 x ) λ = 2 x + y 8 x λ x = 2 x + y 8 From the equation (2), − 8 y + x = λ ( 18 y ) λ = x − 8 y 18 y λ y = x − 8 y 18 Thus, λ x y = y ( 2 x + y ) 8 = x ( x − 8 y ) 18 . Consider y ( 2 x + y ) 8 = x ( x − 8 y ) 18 and solve

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

Publisher: PEARSON

See similar textbooks

Question

Chapter 12.9, Problem 47E

To determine

To find: The absolute maximum and minimum value of the function f(x,y)=x2−4y2+xy on the region 4x2+9y2≤36 .

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 12.9, Problem 46E

Chapter 12.9, Problem 48E

Students have asked these similar questions

Please show all work. Thank you

Find the domain and range and describe the level curves for the function f(x,y). 2 f(x,y)=√100-x² - 2 O A. Domain: all points in the xy-plane Range: real numbers 0 ≤z≤ 10 Level curves: circles with centers at (0,0) and radii r, 0

Consider the function f(x, y) = 3x – 8xy + 6y². Find the maximum and minimum values of f in the region defined by the inequalities -1 < x < 1,0 < y< 1 and where they occur. The maximum value is 11 and it occurs for x = -1 and y = 1 The minimum value is -3 and it occurs for x = -1 and y = 0

Chapter 12 Solutions

Calculus: Early Transcendentals (2nd Edition)

Ch. 12.1 - Give two pieces of information which, taken...Ch. 12.1 - Find a vector normal to the plane 2x 3y + 4z =...Ch. 12.1 - Where does the plane 2x 3y + 4z = 12 intersect...Ch. 12.1 - Give an equation of the plane with a normal vector...Ch. 12.1 - To which coordinate axes are the following...Ch. 12.1 - Describe the graph of x = z2 in 3.Ch. 12.1 - What is a trace of a surface?Ch. 12.1 - What is the name of the surface defined by the...Ch. 12.1 - What is the name of the surface defined by the...Ch. 12.1 - What is the name of the surface defined by the...

Knowledge Booster

The absolute maximum and minimum value of the function f ( x , y ) = x 2 − 4 y 2 + x y on the region 4 x 2 + 9 y 2 ≤ 36 .

Solution Summary: The author calculates the absolute maximum and minimum value of the function f(x,y)=x2-4y

To find: The absolute maximum and minimum value of the function f(x,y)=x2−4y2+xy on the region 4x2+9y2≤36 .

Want to see the full answer?

Chapter 12 Solutions