Lagrang multipliers to find max and min

  • Thread starter naspek
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  • #1
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Homework Statement



The temperature at a point (x, y) on a metal plate is T(x, y) = 4x^2 − 4xy + y^2 .
An ant, walking on the plate, traverses a circle of radius 5 centered at the origin.
Using the method of Lagrange multipliers, find the highest and lowest
temperatures encountered by the ant.


Homework Equations





The Attempt at a Solution



T(x,y) = 4x^2 − 4xy + y^2
gradient of T = (8x - 4y)i + (2y - 4x)j

g(x,y) = x^2 + y^2 = 5^2
gradient of g = (2x)i + (2y)j

gradient of T = (lambda)gradient of g ----> lambda=#

8x - 4y = #2x ---->1
2y - 4x = #2y ---->2

# = 4-4y = 1-4x

what am i going to do next?:confused:
 

Answers and Replies

  • #2
lanedance
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haven't checked your work, but notice you also have the original constraint, use that with the equation that you found
 
  • #3
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haven't checked your work, but notice you also have the original constraint, use that with the equation that you found
what is the meaning of original constraint?
 
  • #4
lanedance
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what is the meaning of original constraint?
you have a function to optimise T(x, y) = 4x^2 − 4xy + y^2, against a given constraint x^2 + y^2 = 5^2
 
  • #5
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if i solved the previous equation, i get, x=(1/4)y
 
  • #6
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i've got max= 45 and min=5..
am i done it right?
 

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  • #7
HallsofIvy
Science Advisor
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Yes, what you have done so far is correct. But that certainly does NOT give "max= 45 and min=5"!
 
  • #8
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ok...
i've checked my calculation already...
when y = 2x,
i substitute into x^2+y^2=25
i got x=+-(5)^1/2

when x=-2y,
i substitute into x^2+y^2=25
i got y= +-(5)^1/2

so.. i got my critical points..
---> [(5^1/2) , (5^1/2)]
---> [(5^1/2) , -(5^1/2)]
---> [-(5^1/2) , (5^1/2)]
---> [-(5^1/2) , -(5^1/2)]

my critical points are correct?:uhh:
 

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