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## 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

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- Thread starter nikk834
- Start date

- #1

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Find the optimal value of the function

f (x,y) = 3.5x^2+y^2-42x-28y+5xy+190

subject to

6x+5y = 37

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

or minimum

- #2

- 674

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- #3

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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?

- #4

ideasrule

Homework Helper

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

- #5

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

- #6

ideasrule

Homework Helper

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http://en.wikipedia.org/wiki/Lagrange_multipliers

Write out the Lagrangian function first, then set its partial derivatives to 0.

- #7

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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?

- #8

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- #9

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- #10

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- #11

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Sure.

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

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

- #12

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- #13

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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?

- #14

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

- #15

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

- #16

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- #17

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

- #18

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

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

- #19

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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

- #20

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dl/dy that is

- #21

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

dL/dL= -(6x+5y-37)=0

So how do you solve for x, y and L

- #22

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- #23

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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?

- #24

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For that, read this:

http://en.wikipedia.org/wiki/Second_partial_derivative_test" [Broken]

http://en.wikipedia.org/wiki/Second_partial_derivative_test" [Broken]

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