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Lagrange Multipliers to find the Maximum and Minimum values

  • Thread starter brendan
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  • #1
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Homework Statement



Use Lagrange Multipliers to find the Maximum and Minimum values of f(x,y) = x2-y.
Subject to the restraint g(x,y) = x2+y2=25

Homework Equations



gradient f(x,y)= gradient g(x,y)

The Attempt at a Solution



I have found the gradients of f and g to be

f(x,y) = 2xi + -1j
g(x,y) = 2xi + 2yj


I have put these to gether with the constraint to find the simutaneous equation

2x = lambda 2x
-1 = lambda 2y
x2+y2 = 25

Have I got this system right so far ?

I'm a bit concerned about the 2x = lambda 2x part.
I'm not sure if this is right or I should have used 2x = lambda 2y ?

In the original system I have Lambda = 1 and y = -1/2

Any help greatly appreciated
Brendan
 

Answers and Replies

  • #2
537
3
I'm a bit concerned about the 2x = lambda 2x part.
I'm not sure if this is right or I should have used 2x = lambda 2y?
Everything looks right to me, and no, you definitely should not have used [itex]2x=\lambda 2y[/itex]. The general system for a function [itex]f(x,y)[/itex] constrained to [itex]g(x,y) = k[/itex] is:
[tex] \begin{align*}
f_x &= \lambda g_x \\
f_y &= \lambda g_y \\
g &= k
\end{align*}[/tex]
So you have done this exactly right. You have found [itex]\lambda=1,y=-1/2[/itex] correctly. What about x?
 
Last edited:
  • #3
Dick
Science Advisor
Homework Helper
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What are you concerned about? I think you are doing fine. 2x = lambda 2y isn't right at all. Why would you think it is?
 
  • #4
65
0
g(x,y) = x2+y2=25

Now we have y = -1/2

so x = sqrt(y2+25) or -sqrt(y2+25)

= sqrt(101)/2 or -sqrt(101)


So we have two solutions
f(x,y)= x2-y = f(sqrt(101)/2,-1/2) = 103/4 = 25.75
f(x,y)= x2-y = f(-sqrt(101)/2,-1/2) = 103/4 = 25.75

So they are both the maximum
 
  • #5
537
3
g(x,y) = x2+y2=25

Now we have y = -1/2

so x = sqrt(y2+25) or -sqrt(y2+25)
Whoops, you have a sign error. It should have been [itex]x = \pm\sqrt{25-y^2}[/itex].
 
  • #6
65
0
g(x,y) = x2+y2=25

Now we have y = -1/2

so x = sqrt(25-y2) or -sqrt(25-y2)

= sqrt(99/4) or -sqrt(99/4)


So we have two solutions
f(x,y)= x2-y = fsqrt(25-y2) ,-1/2) = 101/4= 25.25 (maximum)
f(x,y)= x2-y = f(-sqrt(25-y2) ,-1/2) = -97/4= -24.25 (Minimum)

Is that better?
 

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