Explanation Given: Thefunction to minimize f ( x , y ) = x 2 − 2 y − y 2 subject to x 2 + y 2 = 1 Concept Used: To optimize (i.e., to find the minimum and maximum value of) a function f ( x , y ) subject to the constraint g ( x , y ) = k Solve the following system of equations ∇ f ( x , y ) = λ ∇ g ( x , y ) g ( x , y ) = k where λ is the Lagrange Multiplier. Plug all the solutions ( x , y ) from the above step into f ( x , y ) . The maximum value of the function f ( x , y ) subject to the constraint g ( x , y ) = k is the maximum value of f ( x , y ) after plugging all the solutions ( x , y ) into f ( x , y ) . Calculation: For the givenfunction f ( x , y ) = x 2 − 2 y − y 2 Calculating the first partial derivatives f x and f y of f ( x , y ) f ( x , y ) = x 2 − 2 y − y 2 i.e., f x ( x , y ) = 2 x and, f y ( x , y ) = − 2 − 2 y and, calculating the first partial derivatives g x and g y of g ( x , y ) g ( x , y ) = x 2 + y 2 i.e., g x ( x , y ) = 2 x and, g y ( x , y ) = 2 y Substituting the values in the equation f x ( x , y ) = λ g x ( x , y ) i.e., 2 x = λ ( 2 x ) 2 x − λ ( 2 x ) = 0 ( 2 x ) ( 1 − λ ) = 0 i.e., 2 x = 0 and 1 − λ = 0 i.e., x = 0 - - - - - ( 1 ) and λ = 1 - - - - - ( 2 ) Substituting the values in the equation f y ( x , y ) = λ g y ( x , y ) i.e., − 2 − 2 y = λ ( 2 y ) − 1 − y − λ ( y ) = 0 − 1 − y ( 1 + λ ) = 0 Substituting λ = 1 from equation ( 2 ) i.e., − 1 − y ( 1 + λ ) = 0 − y ( 1 + 1 ) = 1 y = − 1 2 Substituting x = 0 from equation ( 1 ) in constraint x 2 + y 2 = 1 i.e., x 2 + y 2 = 1 0 2 + y 2 = 1 y 2 = 1 i.e., y = − 1 and y = 1 The solutions are ( x , y ) = ( 0 , − 1 ) and ( x , y ) = ( 0 , 1 ) Substituting y = − 1 2 in constraint x 2 + y 2 = 1 i

7th Edition

ISBN: 9781524916817

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

Publisher: Kendall Hunt Publishing

See similar textbooks

Videos

Question

Chapter 11.8, Problem 12PS

To determine

ToMaximize f(x,y)=x2−2y−y2 subject to constraint x2+y2=1 using the method of Lagrange multipliers.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 11.8, Problem 11PS

Chapter 11.8, Problem 13PS

Students have asked these similar questions

This question builds on an earlier problem. The randomized numbers may have changed, but have your work for the previous problem available to help with this one. A 4-centimeter rod is attached at one end to a point A rotating counterclockwise on a wheel of radius 2 cm. The other end B is free to move back and forth along a horizontal bar that goes through the center of the wheel. At time t=0 the rod is situated as in the diagram at the left below. The wheel rotates counterclockwise at 1.5 rev/sec. At some point, the rod will be tangent to the circle as shown in the third picture. A B A B at some instant, the piston will be tangent to the circle (a) Express the x and y coordinates of point A as functions of t: x= 2 cos(3πt) and y= 2 sin(3t) (b) Write a formula for the slope of the tangent line to the circle at the point A at time t seconds: -cot(3πt) sin(3лt) (c) Express the x-coordinate of the right end of the rod at point B as a function of t: 2 cos(3πt) +411- 4 -2 sin (3лt) (d)…

5. [-/1 Points] DETAILS MY NOTES SESSCALCET2 6.5.AE.003. y y= ex² 0 Video Example x EXAMPLE 3 (a) Use the Midpoint Rule with n = 10 to approximate the integral कर L'ex² dx. (b) Give an upper bound for the error involved in this approximation. SOLUTION 8+2 1 L'ex² d (a) Since a = 0, b = 1, and n = 10, the Midpoint Rule gives the following. (Round your answer to six decimal places.) dx Ax[f(0.05) + f(0.15) + ... + f(0.85) + f(0.95)] 0.1 [0.0025 +0.0225 + + e0.0625 + 0.1225 e0.3025 + e0.4225 + e0.2025 + + e0.5625 €0.7225 +0.9025] The figure illustrates this approximation. (b) Since f(x) = ex², we have f'(x) = 0 ≤ f'(x) = < 6e. ASK YOUR TEACHER and f'(x) = Also, since 0 ≤ x ≤ 1 we have x² ≤ and so Taking K = 6e, a = 0, b = 1, and n = 10 in the error estimate, we see that an upper bound for the error is as follows. (Round your final answer to five decimal places.) 6e(1)3 e 24( = ≈

2. [-/1 Points] DETAILS MY NOTES SESSCALCET2 6.5.015. Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n. (Round your answers to six decimal places.) ASK YOUR TEACHER 3 1 3 + dy, n = 6 (a) the Trapezoidal Rule (b) the Midpoint Rule (c) Simpson's Rule Need Help? Read It Watch It

Chapter 11 Solutions

Calculus

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.

ToMaximize f ( x , y ) = x 2 − 2 y − y 2 subject to constraint x 2 + y 2 = 1 using the method of Lagrange multipliers.

Solution Summary: The author explains the Lagrange Multiplier's method of optimizing a function by plugging all the solutions into f(x,y)

Videos

ToMaximize f(x,y)=x2−2y−y2 subject to constraint x2+y2=1 using the method of Lagrange multipliers.

Want to see the full answer?

Chapter 11 Solutions