# Lagrange Function

## Homework Statement

Find the optimal value of the function
f (x,y) = 3.5x^2+y^2-42x-28y+5xy+190
subject to
6x+5y = 37

## Homework Equations

Use the second order condition to determine if the optimal point is maximum
or minimum

## The Attempt at a Solution

You need to at least attempt a solution if you want help. You can start by using Lagrange multipliers.

df/dx=12.25x -42 +5y
df/dy=2y-28+5x
df/dlagrange = -(6x+5y-37)

You make them all equal to 0 so
df/dx=12.25x+5y-42=0
df/dy=5x+2y-28=0
df/dlangrange=6x+5y-37=0
Is this right? then you find a common factor between the first two equations in order to solve for x and y? how do you find lagrange?

ideasrule
Homework Helper
No, that's not right. What's the Lagrangian function?

I'm not sure. I am having difficulty solving this equation.

dL/dx= 12.25x-42+5y-lagrange
dL/dy= 5x-28+2y-lagrange
dL/dL=-(6x+5y-37)

So 12.25x-42+5y-lagrange=0
5x-28+2y-lagrange=0
-(6x+5y-37)=0

What do you do next?
Multiple them by a common factor.what do you do with lagrange?

Why did you just type out the word "lagrange"? As ideasrule said before, first write out the lagrange function before you start taking derivatives.

I typed out the word lagrange because i do not have the backward L function on my keyboard. It means -backward l . I know how to get the derivative now . I just don't know how to solve for x, y and lagrange(backward l)

You solve for the lagrange function first then take the derivative. Can we at least see your lagrange function first.

Sure.
The function would be:f (x,y) = 3.5x^2+y^2-42x-28y+5xy+190 - backward L(6x+5y-37)

You should definitely learn to use tex, but you can just use "L" for lambda and leave out "backward". That looks fine to me. So now you can take the partial derivatives, but this time use "L" and not "lagrange" since that is hard to understand.

Ya so the derivative are:
dL/dx= 12.25x-42+5y-L
dL/dy= 5x-28+2y-L
dL/dL=-(6x+5y-37)

You then set them to 0 so they are
dL/dx= 12.25x-42+5y-L=0
dL/dy= 5x-28+2y-L=0
dL/dL=-(6x+5y-37)=0
Then, could you times the 2nd equation there which is dL/dy by 2 to give you common y terms of 10y?

What happened to the terms multiplied by L? There was a (6x+5y-37) multiplied by L.

so your saying you would put the dl/dl as -L(6X+5Y-37)=0

I am going by an example that the teacher gave us and he said that the dl/dl would just go as -(6x+5y-37)

i just do not know how to solve for x, y and l

No, the first 2 partial derivatives you did. There are x and y terms multiplied by the L. What happened to them? For example:

dL/dx = 12.25x-42+5y-6L = 0

It should be 6L, because you have a 6x.

I see what you mean now.
So there is a 5 in front of the y for the constraint so that means that
dy/dl would have to be : 5x-28 +2y-5L=0

dl/dy that is

dL/dx = 12.25x-42+5y-6L = 0

dL/dy= 5x-28 +2y-5L=0

dL/dL= -(6x+5y-37)=0
So how do you solve for x, y and L

Well you have 3 equations and 3 unknowns, you just work through the algebra to solve for x and y. You can use substitution or any method you prefer.

ok when i solve x, y and lagrange they want me to find this
Use the second order condition to determine if the optimal point is maximum
or minimum.
that means i take the second derivative and how do i determine if it is a max or min?