## anonymous 5 years ago Use the Lagrange multiplier technique to find the maximum and minimum values of on the ellipse defined by the equations and , and where they occur

1. anonymous

what are the equations?

2. anonymous

what is the function more importantly.

3. anonymous

f(x,y,z)=x+z^2 g(x,y,z)=x^2+y^2 h(x,y,z)=2y+z

4. anonymous

Maximize f(x,y,z) subject to the restriction g and h?

5. anonymous

actually i know how to solve

6. anonymous

but are these g and h related or both different

7. anonymous

yeah maximize with restriction g and h i think

8. anonymous

I would let G(x,y,z) = ( x^2 + y^2 , 2y + z) where this our constrained set.

9. anonymous

so now partial derivative of f(x)=lamba(g(x))

10. anonymous

yes,

11. anonymous

$\frac{ part( f}{par( x,y,z)} = (\lambda_1, \lambda_2, \lambda_3)\frac{par G}{par (x,y,z)}$

12. anonymous

if its two dimensional we need to take only 1 lambda then why 3 for 3D?

13. anonymous

I don't know rigorous you're calc 3 is. Do you need to prove that G is a compact set? and f is continuous, then by Extreme Value Theorem our max actually exists.

14. anonymous

I don't know how you need 1 eigenvalue for 2. You need 2.

15. anonymous

lambda was probably written as a vector.

16. anonymous

17. anonymous

basically if you have 2D then partial f(x,y)=partial (lambda (g,y)) then slove for lambda and find x and y then we can get max or min

18. anonymous

but here its 3D and i dont know how to solve this

19. anonymous

20. anonymous

your part(g)/part(x,yz) is going to be a matrix 3x3.

21. anonymous

find its eigenvalues using characteristic polynomial

22. anonymous

or however you find eigenvalues.

23. anonymous

but here f(x,y,z)=lambda g(x,y,z)or f(x,y,z)=lambdah(x,y,z) how i beed to proceed

24. anonymous

Yea so what we did was COMBINE our restrictions into one function lets say W(f(x,y,z), h(x,y,z)) Thats what I meant by capital G up there.

25. anonymous

W(x,y,z) = ( x^2 + y^2 , 2y + z)

26. anonymous

| 2x 2y 0 | | 0 2y 1 | | |

27. anonymous

turns out to be our Part(W)/dx,y,z. We only need 2 lambdas in this case. The eigenvalues depend on the rank of our matrix.

28. anonymous

even a derivative is technically a matrix, so you were right when you said you only needed one lambda

29. anonymous

so we have par f(x,y,z) = (lambda1, lambda2) | 2x 2y 0 | | 0 2y 1 | is your linear algebra good enough to find lambdas?

30. anonymous

ok i will try to solve and wil you be available online??

31. anonymous

what do you get as partial f(x,y,z)

32. anonymous

<1,0,2z>

33. anonymous

<1,0,2z> = <lam1, lam2> | 2x 2y 0 | | 0 2y 1 |

34. anonymous

multiply the lambda through the matrix. then you should be able to pick out a linear system of equations. Use that to find lam1 and lam2

35. anonymous

here 3 equations and 5 unknowns

36. anonymous

Don't worry we are just trying to find expressions for x,y,z not explicit answers. what did you get?

37. anonymous

i got lam1=1/2x 2y(lam1+lam2)=0 lam2=2z

38. anonymous

try to get expressions for x, y,z and remember you can still use you're functions g, h

39. anonymous

BTW, are you sure the functions weren't equal to anything?

40. anonymous

no

41. anonymous

g(x,y,z)=x^2+y^2 h(x,y,z)=2y+z These two can describe tons of functions.

42. anonymous

x^2 + y^2 = 5, is certainly not the same as x^2 + y^2 = 10, and they will have max's in different points.

43. anonymous

if i solve i got 2 points (1,0,0) and (-1,0,0) are these correct??

44. anonymous

or are we doing a general case? set them equal to x^2 + y^2 = a 2y+z = b. How did you do that..

45. anonymous

i solved the above equations with g(x) ,h(x)

46. anonymous

x^2 + y^2 = 1 and 2y+z = 0 with the equations with lambdas

47. anonymous

oh so they were equal to something.

48. anonymous

COOL! i was getting scared.

49. anonymous

am i doing correct or the points i got correct and the functional values at this point gives the value??

50. anonymous

I hope so, don't make me do the algebra =P

51. anonymous

x^2 + y^2 = 1 and 2y+z = 0 was this given?

52. anonymous

yes

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