Lagrange Multipliers to find the Maximum and Minimum values

In summary, Brendan found two solutions to the homework equation f(x,y) = x2-y. The maximum value is 25.75 and the minimum value is -24.25.
  • #1
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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  • #2
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*}
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
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
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
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].
 
  • #6
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?
 

1. What is the concept behind Lagrange Multipliers?

The concept of Lagrange Multipliers is based on the idea of finding the maximum or minimum value of a function subject to a set of constraints. It involves using a mathematical technique to find the points where the gradient of the function is parallel to the gradient of the constraint.

2. How do Lagrange Multipliers work?

Lagrange Multipliers work by introducing a new variable, called a Lagrange multiplier, into the original function and then setting the partial derivatives of the new function equal to zero. This results in a system of equations that can be solved to find the maximum or minimum value of the function subject to the given constraints.

3. When is it necessary to use Lagrange Multipliers?

Lagrange Multipliers are necessary to use when optimizing a function subject to a set of constraints where the constraints cannot be easily solved for one of the variables. It is also useful when the constraint is an inequality, as it allows for the determination of the maximum or minimum value within that constraint.

4. Can Lagrange Multipliers be used for both single and multi-variable functions?

Yes, Lagrange Multipliers can be used for both single and multi-variable functions. For single variable functions, the constraint is typically an inequality, while for multi-variable functions, there can be multiple constraints, each with their own Lagrange multiplier.

5. Are there any limitations to using Lagrange Multipliers?

One limitation of Lagrange Multipliers is that it only works for continuous functions. Additionally, it may not always guarantee the global maximum or minimum value, as it only finds the critical points of the function within the given constraints.

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