Question

Hi ,

i need help with Lagrange multipliers the question is as follows can you please go though it step by step any help is much appreciated thank you :) .


Minimise

p1+p2=60

Constraints
1-p1+p1^2=c1
1-0.6p2+p2^2=c2

Answer 1

I don't know how to approach this freaking question ? can anyone outline the steps i needs to find the minimum cost?

Answer 2

Just goggled multiplier formula:

i guess:

l(p,lamda)= p1+p2-60-l1(1+0.6p2+p2^2)-l2(1-p1+p2^2)

is this correct?

Answer 3

Grrr I can't remember how to do these lol.
Lemme write some stuff down and maybe it will jog my memory.

Answer 4

okay cool im new to this stuff...

Answer 5

your equation should be
\[ f(p_1, p_2, L_1, L_2)= p_1 + p_2-60+ L_1(1-p_1+p_1^2-c_1) +L_2(1-0.6p_2+p_2^2-c2)\]

Answer 6

I assume p1 and p2 are variables.

now take the 4 partial derivatives of f

Answer 7

right so the first part i did kinda made sense...

Answer 8

Yest p1 and p2 must be variables i guess , what about the l1 and l2 values should i expand the brackets ?

Answer 9

it doesn't matter much.

what do you get for the d f/ dp1 ?

Answer 10

that l1 and l2 is confusng me ?

Answer 11

you treat them like variables p1 and p2

when you take the derivative of f() with respect to p1, all variables except p1 are treated as a constant.

Answer 12

1+l2=0 l2=-1 ? df/p1

Answer 13

among the terms
there is the term L1(1-p1+p1^2- c1)
d /d p1 = -L1 + 2 p1

Answer 14

lol i missed i 2pi i suck .. so next i have to do p2 i guess?

Answer 15

yes, the df/dL1 and df/dL2

Answer 16

so df/p2 =1-0.6l2 +2p2 ?

Answer 17

derivative df/dp2
f()= p1+p2−60+L1(1−p1+p1^2−c1)+L2(1−0.6p2+p2^2−c2)
1 -0.6*L2 + 2* p2
yes, that looks good

Answer 18

cool let me try to work out l1 and l2 quicky

Answer 19

hmmm do i have to expand the brackets for l1 and l2?

Answer 20

no, just treat the (stuff) in L1*(stuff) as a constant c*L1 and derivative is c

Answer 21

df/dl1=1-p1+p1^2
df/dl2= 1-0.6p2+p2^2 ?

Answer 22

for df/d L1 we have only this term with L1 :
L1(1−p1+p1^2−c1)

you get
(1−p1+p1^2−c1)

Answer 23

so i guess df/dl2= (1-0.6p2+p2^2-c2) i expanded it lol should have left it factorized ...

Answer 24

So i think you have to equal them to zero to find the stationary points ?

Answer 25

yes, and that is often the hardest part. You often seem to get ugly equations using Lagrange...

Answer 26

so for df/dp1 do imake p1 the subject ?

Answer 27

i have never done or approached Lagrange ever before completely new to this...

Answer 28

at this point you have 4 equations and 4 unknowns. But they are non-linear equations, which makes them harder to solve.

once you solve for all four "variables", you plug them into the original equation to get the optimum f() value (subject to the constraints)

I have not solved these yet...

Answer 29

like this df/dp1=df/dp2 and df/dl1= df/dl2 ?

Answer 30

since its =0 i can equate them cant i?

Answer 31

I think you can solve for L1= -1/(2p1-1)
L2= -1/(2 p2 - 0.6)

use the quadratic formula to solve for p1 from the 3rd equation
same for p2 from the 4th equation.

put all of that mess into the original

Answer 32

df/dl1=1+root3/2i and 1root3/2i

df/dl2= 3/10 +root91/10i and 3/10 -root91/10i ?

Answer 33

sorry not df ...

Answer 34

1−p1+p1^2−c1=0

I assume that c1 is an unknown constant

Answer 35

yes c1 and c2

Answer 36

subing it back in now...

Answer 37

these can't be correct
df/dl1=1+root3/2i and 1root3/2i

df/dl2= 3/10 +root91/10i and 3/10 -root91/10i ?

because we have c1 and c2 in the originals equations

Answer 38

df/dl1=0

c1=1-pi+p1^2= 1+root3/2i and - 1root3/2i ? and so on ...

Answer 39

no,

you have 1−p1+p1^2−c1=0
p1^2 - p1 + (1-c1) = 0

match to ax^2 + bx + c= 0
a=1, b= -1 , c= (1-c1)

Answer 40

ah ok ...

Answer 41

=1+or - root(-1)^2-4*1*(1-xc1)/2(1-c1)

Answer 42

its seems to be unsolvable..

Answer 43

we won't get a nice number, but a messy expression in terms of c1 and c2

Answer 44

so you needs to rearrange to get c1 and c2 the qudratic formula?

Answer 45

is this correct =1+or - root((-1)^2-4*1*(1-xc1))/2(1-c1) how do i proceed from here? for c1

Answer 46

the solution to
\[ p_1^2 - p_1 + (1-c1) = 0 \]
is
\[ p_1= \frac{1 \pm\sqrt{4c_1-3}}{2} \]

Answer 47

damn stupid mistake...1 i see!

Answer 48

p2=1+ - root(-0.6)^2-4*(1-c2)/2

p2=1+ - 0.6-*2root(1-c2)/2

Answer 49

canel the 2 also

Answer 50

\[ p2= \frac{1}{2} (1 \pm \sqrt{0.36 -4 +4c_2} )\]

Answer 51

right now sub into the original equation?

Answer 52

I used SAGE (math program that does symbolic computation) to find
\[ \frac{1}{10}\sqrt{100 c_2 - 91} + \frac{1}{2}\sqrt{4 c_1 - 3} - \frac{296}{5} \]

Answer 53

yes, first sub these into L1 = equation
and L2 = equation so both L1 and L2 are functions of c1 and c2
put those into the original

replace p1 and p2 with the values just found. It is very messy.

Answer 54

ok so how do you find the minimum cost from this ?

Answer 55

by simplifying the original equation after subbing in for p1, p2, L1, and L2
you get what I posted

Answer 56

so what you posed is the minimum cost ... i thought it will be a number or something ...

Answer 57

your constraints are
Constraints
1-p1+p1^2=c1
1-0.6p2+p2^2=c2

so the answer will be in terms of c1 and c2

if you have numbers for c1 and c2, then you can get a number

Answer 58

ah i see can i post the original question to see if you have done it correct ?

Answer 59

yes

Answer 60

Two generators are connected to provide total power to meet load demands. The costs of the two generators are each expressed as functions of power output, C1=1-P1+p1^2 and C2 =1 +O.6P2+P2^2


(i) . If the load demand requires at least 60' units, formulate a minimum cost design problem {cost' and power are on a per unit basis}, identifying design variables, the objective, and all constraints.

(ii) By introducing appropriate slack varlable(s), use Lagrange multipliers to find the minimum cost power distribution solution.




(iii) Sketch a graphical interpretation of the optimization problem.