# Lagrange Multipliers

1. Apr 4, 2009

### brendan

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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

2. Apr 4, 2009

### n!kofeyn

Everything looks right to me, and no, you definitely should not have used $2x=\lambda 2y$. The general system for a function $f(x,y)$ constrained to $g(x,y) = k$ is:
\begin{align*} f_x &= \lambda g_x \\ f_y &= \lambda g_y \\ g &= k \end{align*}
So you have done this exactly right. You have found $\lambda=1,y=-1/2$ correctly. What about x?

Last edited: Apr 4, 2009
3. Apr 4, 2009

### Dick

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. Apr 4, 2009

### brendan

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. Apr 4, 2009

### n!kofeyn

Whoops, you have a sign error. It should have been $x = \pm\sqrt{25-y^2}$.

6. Apr 5, 2009

### brendan

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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