Question

Chapter 13, Section 13.9, Question 006 Consider the function f (x, y) = 1x2 – 5y2 subject to the condition x² + y2 = 9. Use Lagrange multipliers to find the maximum and minimum values of f subject to the constraint. Maximum: Minimum: Find the points at which those extreme values occur. (3,0), (0,3), and (3,3) O (-3,0) and (0, – 3) (3,0), (-3,0), (0,3), and (0, – 3) O (3,0), (-3,0), (0,3), (0, – 3), (3,3), and (-3, - 3) O (3,0) and (0,3)

Answer 1

Using Lagrange multipliers method We find out the maximum value and minimum value of the given function.

We find out the point that closer to the origin by using Lagrange multipliers method.
Given function teniu)= n²-5yr subject to the condition xty=9. Le+ * 2) = ****- 9 Now wring tag mange multipliers method ofency) = xx g(ny) 5 6 7:52 он 7 2020 кL - t'xt >:((47) s -* ки 7 е 2 І. (2x , - 2017 = K271, 2 ч) 2х - ХХ and - Loy = 2xy * 5 ++) - 22м (1-4)=0 2 oor woulle Then artyrag 9 a Ot 9-9 yrag Я = + 3 When och Then xtyrag ЭцY+ o-9 аҳ9 • Ус + 2 Therefore critical points are (3,0), (-30), (0,3) and (0-3). А (9,9), f( 3 °) = 3- 5(9) Y = 9

At (-3,0), f(-3,0) =(-3)25COV=9 At(03), 7(0,3) > 09-5137)= - 45 +40,-3) = 0^-51-3)*= -45 At (0,-3), So Mornimum : 9 Minimum : -45 The points at which extrema value occur. (340),(-3,0) 0,3) and (0,-3) 17 Let npoint 180 on Now (nill be the the line 6n. 61-12 yols that is closed to the onigin. Distance between (2,4) to (0,0) is feny) = dra x tyr Again let geniye bh-ly-180 method using Lagrange multipliers afcx,y)= igenly) where x is Lagrange multiplier. of an, of any) = x < 0 per ty a < 2x, ry> -, ry>=x6,-12) » <2n, 2 <6X127) 2n=67, 242-129 7 ,24) ..

44 г. е and (9 -- Now hoi-ng | И Э 6 (35) --62) = 1 - | И Г 2 18, 17,2 10 9.4 2 1970 А2 2 So та а ? з (2) = 6 and • -6(2) = - 12 Therefore the critical point is (6,12) AF (6-12) f(",-1) = (6 )** + (-72 V 2 34 + 144 ТВО 65 When Again och So d=J180 Then 67 - 12 (9) = аби 2 (20 : Зо Тkor 30,9) Цири the line см-123 = 120 каа 30, ) = (30)"tov, 900 so da Jooo Therefore 20 d=685 is minemum for (19)= ( 6r12) (яу) = (6], E12 9