Lagrange Multipliers to find the Maximum and Minimum values

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Homework Help Overview

The discussion revolves around using Lagrange Multipliers to find the maximum and minimum values of the function f(x,y) = x² - y, subject to the constraint g(x,y) = x² + y² = 25.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to set up the system of equations derived from the gradients of f and g, expressing concern about the correctness of their equations. Some participants affirm the setup while others clarify the correct relationships between the variables.

Discussion Status

The discussion is active, with participants providing feedback on the original poster's equations and suggesting corrections. There is a mix of agreement on certain steps and questioning of assumptions, particularly regarding the relationships between x, y, and λ.

Contextual Notes

Participants note potential sign errors and the importance of correctly applying the constraint in the context of the problem. There is an ongoing exploration of the implications of the constraint on the values of x and y.

brendan
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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
 
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brendan said:
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*}<br /> f_x &= \lambda g_x \\<br /> f_y &= \lambda g_y \\<br /> g &= k<br /> \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:
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?
 
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
 
brendan said:
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].
 
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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