Named after Joseph Louis Lagrange.Lagrange multipliers provides a strategy for

finding the maxima and minima of a function subject to constraints.

This method gives a set of necessary conditions to identify optimal points of equality constrained optimization problems.

Done by converting a constrained problem to an equivalent unconstrained problem with the help of certain unspecified parameters known as Lagrange multipliers.

One of the most common problems in calculus is that of finding maxima or minima of a function.

Difficulties often arise when one wishes to maximize or minimize a function subject to fixed outside conditions or constraints.

The method of Lagrange multipliers is a powerful tool for solving this class of problems.

Original problem is rewritten as: minimize L(x, λ) = f(x) - λ h1(x) .

Take derivatives of L(x, λ) with respect to xi and set them equal to zero.

(If there are n variables (i.e., x 1, ..., xn) then you will get n equations with n + 1 unknowns (i.e., n variables xi and one Lagrangian multiplier λ)

Express all xi in terms of Langrangian multiplier λ

Plug x in terms of λ in constraint h1(x) = 0 and solve λ.

Calculate x by using the just found value for λ. The constant, , is called the Lagrange Multiplier.                              

Here, f(x, y) = 5x – 3y constraint g(x, y) = x² + y²-136 Thus, δf(x, y)/δx = 5 δ f(x, y)/δy = -3 δg(x, y)/δx = 2 x δg(x, y)/δy = 2 y

• Now, applying δf(x, y)/δx = λ δg(x, y)/δx δf(x, y)/δy = λ δg(x, y)/δy We get following set of equations .. 5 = 2 λx -3 = 2 λy x² + y²=136 (constraint) Solving these equations x = 5/2 λ y = -3/2 λ Plugging these into the constraint we get λ = ± ¼ If λ = ¼ then, x=10 and y=-6 if λ=-1/4 then, x=-10 and y= 6

To determine if we have maximums or minimums we just need to plug these into the function.

Here are the minimum and maximum values of the function. f(-10,6) = -68 Minimum at(-10,6) f(10,-6) = 68 Maximum at(10,-6)

Here, f(x, y, z) = xyz (volume of the box) constraint 2(x y + y z + z x) = 64 (total surface

area) i.e. x y + y z + z x = 32 g(x ,y ,z) = x y + y z + z x-32 Thus, we get.. δf(x, y, z)/δx = y z δf(x, y, z)/δy = x z δf(x, y, z)/δz = y x δg(x, y, z)/δx = (y + z) δg(x, y, z)/δy = (x + z) δg(x, y, z)/δz = (y + x)

Now, applying δf(x, y, z)/δx = λ δg(x, y, z)/δx δf(x, y, z)/δy = λ δg(x, y, z)/δy δf(x, y, z)/δz = λ δg(x, y, z)/δz We get following set of equations .. y z = λ (y + z) x z = λ (x + z) y x = λ (y + x) x y + y z + z x = 32 (constraint) Applying either elimination or substitution method we

now solve the set of equations thus obtained.. Thus, x = y = z = 3.266

We can say that we will get a maximum volume if the dimensions are

x = y = z = 3.266

Find the maximum and minimum values of f(x, y, z) = x y z subject to the constraint x + y + z = 1.  Assume that x, y, z ≥ 0.

  Find the maximum and minimum values of f(x, y) = 4x² + 10y² on the disk while x² + y²≤ 4.

Consider the following…….Find the maximum and minimum of f(x, y, z) = 4y

– 2z subject to constraints 2x – y – z = 2 and x² + y² =

1.

If g1=0, g2=0, g3=0, ………, g n=0 are n number of constraints then..

Solve the following system of equations.         f(P) =  λ1 g1(P) + λ2  g2(P) + … + λn   g n(P) g1(P) = k1 g2(P) = k2 . . . g n(P) = k n• Plug in all solutions, (x, y, z),

from the first step into f(x, y, z) and identify the minimum and maximum values, provided they exist.

• The constants, λ1 , λ2, ….., λn is called the Lagrange Multiplier.

                          

Here, f(x1,x2,x3,x4,x5)= x1

2+x22+x3

2+x42+x5

2

Constraints g1(x1,x2,x3,x4,x5)=x1+2x2+x3-1 g2(x1,x2,x3,x4,x5)=x3-2x4+x5-6Thus, δf(x1,x2,x3,x4,x5)/δx1 =2x1 δf(x1,x2,x3,x4,x5)/δx2 =2x2

δf(x1,x2,x3,x4,x5)/δx3 =2x3

δf(x1,x2,x3,x4,x5)/δx4 =2x4

δf(x1,x2,x3,x4,x5)/δx5 =2x5

δg1(x1,x2,x3,x4,x5)/δ x1 =1 δg1 (x1,x2,x3,x4,x5)/δx2 =2δg1 (x1,x2,x3,x4,x5)/δx3 =1δg1 (x1,x2,x3,x4,x5)/δx4 =0δg1 (x1,x2,x3,x4,x5)/δx5 =0δg2(x1,x2,x3,x4,x5)/δ x1 =0 δg2 (x1,x2,x3,x4,x5)/δx2 =0δg2 (x1,x2,x3,x4,x5)/δx3 =1δg2 (x1,x2,x3,x4,x5)/δx4 =-2δg2 (x1,x2,x3,x4,x5)/δx5 =1On applying, δf(x1,x2,x3,x4,x5)/δx1 = λ δg1(x1,x2,x3,x4,x5)/δ x1 + µ δg2(x1,x2,x3,x4,x5)/δ x1

δf(x1,x2,x3,x4,x5)/δx2= λ δg1 (x1,x2,x3,x4,x5)/δx2 + µ δg2 (x1,x2,x3,x4,x5)/δx2

δf(x1,x2,x3,x4,x5)/δx3= λ δg1(x1,x2,x3,x4,x5)/δ x3 + µ δg2(x1,x2,x3,x4,x5)/δ x3

δf(x1,x2,x3,x4,x5)/δx4= λ δg1(x1,x2,x3,x4,x5)/δ x4 + µ δg2(x1,x2,x3,x4,x5)/δ x4

δf(x1,x2,x3,x4,x5)/δx5= λ δg1(x1,x2,x3,x4,x5)/δ x5 + µ δg2(x1,x2,x3,x4,x5)/δ x5We get following set of equations .. 2x1 + λ=0 , 2x2 +2 λ=0 , 2x3 + λ + µ =0 , 2x4 -2 µ =0 , 2x5 + µ =0 x1+2x2+x3=1 , x3-2x4+x5=6 (constraint)

On Solving these equationsµ=-2, λ=0 Plugging these into the equations we get x1=x2=0, x3=x5=1 and x4=-2To determine minimums we just need to plug these into the function.Here ,the minimum values of the function. f(0,0,1,-2,1)=6.

Here, f(x , y , z)=4y-2zConstraint, g1(x ,y ,z)=2x-y-z-2 g2(x ,y ,z)= x² + y² -1Thus, δf(x ,y ,z)/δx =0 δf(x ,y ,z)/δy =4 δf(x ,y ,z)/δz =-2 δg1(x ,y ,z)/δx =2 δg1(x ,y ,z)/δy =-1

δg1(x ,y ,z)/δz =-1 δg2(x ,y ,z)/δx=2x δg2(x ,y ,z)/δy =2y δg2(x ,y ,z)/δz=0On applying, δf(x ,y ,z)/δx= λ δg1 (x ,y ,z)/δx + µ δg2 (x ,y ,z)/δx δf(x ,y ,z)/δy= λ δg1 (x ,y ,z)/δy+ µ δg2 (x ,y ,z)/δy δf(x ,y ,z)/δz= λ δg1 (x ,y ,z)/δz + µ δg2 (x ,y ,z)/δzWe get following set of equations .. 2 λ+2x µ=0, - λ+2y µ=4, - λ=-2, 2x – y – z = 2, (constraint) x² + y² = 1 (constraint)

On Solving these equations…λ=2, µ=+5,-5 Plugging these into the equations we getIf µ=+5 ,then x=0.8,y=-0.6,z=0.2 and If µ=-5 ,then x=-0.8,y=0.6,z=-4.2To determine if we have maximums or minimums

we just need to plug these into the function.

Here are the minimum and maximum values of the function.

f(0.8,-0.6,0.2)=-2.8 minimum at(0.8,-0.6,0.2)

f(-0.8,0.6,-4.2)=10.8 maximum at(-0.8,0.6,-4.2)

Lagrangian multipliers are very useful in sensitivity analyses.

Setting the derivatives of L to zero may result in finding a saddle point. Additional checks are always useful.

Lagrangian multipliers require equalities. So a conversion of inequalities is necessary.

Kuhn and Tucker extended the Lagrangian theory toinclude the general classical single-objective nonlinearprogramming problem:

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