Lagrange Multipliers to find max/min values


by arl146
Tags: lagrange, max or min, multipliers, values
arl146
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#1
Oct25-11, 03:55 PM
P: 343
1. The problem statement, all variables and given/known data
Use Lagange Multipliers to find the max and min values of the function subject to the given constraint(s). f(x,y)=exp(xy) ; constraint: x^3 + y^3 = 16


2. Relevant equations
[itex]\nabla[/itex]f = [itex]\nabla[/itex]g * [itex]\lambda[/itex]
fx = gx * [itex]\lambda[/itex]
fy = gy * [itex]\lambda[/itex]


3. The attempt at a solution
Set the fx and fy eqns equal to 0. but i cant solve for x, y, and lambda... i guess my algebra isnt that strong

i got fx = [itex]\lambda[/itex] * gx
y*exy = 3x2[itex]\lambda[/itex]

and for y:

x*exy = 3y2[itex]\lambda[/itex]

and g(x,y) = x3 + y3 = 16
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arl146
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#2
Oct25-11, 05:32 PM
P: 343
no one can offer a little help? =/
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#3
Oct25-11, 06:26 PM
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Hi arl146!

You need to solve this set of equations:
(1) y exy = 3x2λ
(2) x exy = 3y2λ
(3) x3 + y3 = 16

Can you find λ from equation (1)?
And also from equation (2)?
Then equate them to each other, effectively eliminating λ?

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#4
Oct25-11, 07:41 PM
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Lagrange Multipliers to find max/min values


Since a specific value for [itex]\lambda[/itex] is not necessary for the solution, I find it is often simplest to start by eliminating [itex]\lambda[/itex] by dividing one equation by another. Here, start by dividing [itex]ye^{xy}= 3x^2\lambda[/itex] by [itex]xe^{xy}= 3y^2\lambda[/itex]: [itex]y/x= x^2/y^2[/itex] which is the same as [itex]x^3= y^3[/itex]. Putting that into [itex]x^3+ y^3= 16[/itex] gives [itex]2x^3= 16[/itex].
arl146
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#5
Oct26-11, 12:52 AM
P: 343
well i sorta did something like that and got and y=x and with the whole x^3 + y^3 = 16 that means y=x=2. so theres what, is it called a critical pt still, at (2,2) ? so i just plug that into f(x,y)= exp(xy) ??
so you'd have f(2,2)=exp(4) .... is that all i do ? it just seems wrong.idk why haha
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#6
Oct26-11, 01:41 AM
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Yes that's all you do. :)

The only thing remaining is finding out whether it's a maximum or a minimum...
arl146
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#7
Oct26-11, 01:58 AM
P: 343
yea i think thats why i came here because i got confused with that one value. how do you know?
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#8
Oct26-11, 04:24 AM
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Quote Quote by arl146 View Post
yea i think thats why i came here because i got confused with that one value. how do you know?
Doesn't your class material cover that?

Anyway, I know of 3 methods:

1. Using the second derivative test (Hessian matrix).
I can't quickly find a easy example for it (yet).

2. Since you only have one extrema, you can pick any point that satisfies the constraint and calculate f(x,y) there. Compare it with the f(x,y) at the extremum and you know whether it's a maximum or a minimum.

3. Vary x with a small epsilon, and calculate how you need to vary y to match the constraint in first order approximation.
In your case (x + epsilon)^3 + (y - epsilon)^3 = 16.
Check what f(x + epsilon, y - epsilon) does relative to f(x,y).


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