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function is f (X,y) = 400yz and the unit sphere is x + y +z? = 1 We have to find the highest temperature on the unit sphere 8 (x,y)= x' +y +z² = 1 Lagrange's multiplier's condition is Vf = AVg af af af Əx' Əy' dz |= (400 yz",400xz",800xyz) Now Vf = ag ag ög Vg= = (2x,2y, 2z) Əx' Əy' dz

function is f (X,y) = 400yz and the unit sphere is x + y +z? = 1 We have to find the highest temperature on the unit sphere 8 (x,y)= x' +y +z² = 1 Lagrange's multiplier's condition is Vf = AVg af af af Əx' Əy' dz |= (400 yz",400xz",800xyz) Now Vf = ag ag ög Vg= = (2x,2y, 2z) Əx' Əy' dz

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Given function is f (x,y) = 400xyz? and the unit sphere is x+y +z? =1 We have to find the highest temperature on the unit sphere 8(x,y)= x +y° +z² = 1 Lagrange's multiplier's condition is Vf = AVg J=(400,x²,400z",800-yz) Now Vf = ag ag ög Vg = Əx' ay' az |= (2x, 2y, 2z) : Vf = AVg = (400 yz", 400xz",800 xyz) = 1 (2x, 2y, 2z) →400yz = 22x 400xz = 24y ... (1) .(2) 800xyz = 2Az .(3) x² +y? +z? = 1 ... (4) Solving these 4 equati ons, we get A = 0,±100 Substituting A = 0 in (1) ,(2) and (3), we get the following points :(-1,0, 0).(0.-1,0).(0,0,–1).(1,0,0) (0,1,0). (0,0,1) Substituting A =-100 in (1) ,(2) and (3), we get the 1 1 1 2'2 2, following points 1 Substituting A = 100 in (1) .(2) and (3), we get the 1 1 1 following points 1 1 . f has possible extreme values at the points (-1,0, 0).(0,–1,0).(0,0,–1).(1,0,0).(0,1, 0). (0,0,1), 1 1 1 1 1
Transcribed Image Text:Given function is f (x,y) = 400xyz? and the unit sphere is x+y +z? =1 We have to find the highest temperature on the unit sphere 8(x,y)= x +y° +z² = 1 Lagrange's multiplier's condition is Vf = AVg J=(400,x²,400z",800-yz) Now Vf = ag ag ög Vg = Əx' ay' az |= (2x, 2y, 2z) : Vf = AVg = (400 yz", 400xz",800 xyz) = 1 (2x, 2y, 2z) →400yz = 22x 400xz = 24y ... (1) .(2) 800xyz = 2Az .(3) x² +y? +z? = 1 ... (4) Solving these 4 equati ons, we get A = 0,±100 Substituting A = 0 in (1) ,(2) and (3), we get the following points :(-1,0, 0).(0.-1,0).(0,0,–1).(1,0,0) (0,1,0). (0,0,1) Substituting A =-100 in (1) ,(2) and (3), we get the 1 1 1 2'2 2, following points 1 Substituting A = 100 in (1) .(2) and (3), we get the 1 1 1 following points 1 1 . f has possible extreme values at the points (-1,0, 0).(0,–1,0).(0,0,–1).(1,0,0).(0,1, 0). (0,0,1), 1 1 1 1 1
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