The Lagrange multiplier method provides a strategy for finding the maxima and minima of a function subject to constraints. It involves setting up a system of equations involving the function, its derivatives, and the constraints and their derivatives. Solving this system of equations yields candidate maxima/minima points, which are then checked in the original function to determine if they are actually maxima or minima. The document provides examples of applying the Lagrange multiplier method to problems with single and multiple constraints.

2.  Named after Joseph Louis Lagrange.
 Lagrange multipliers provides a strategy for finding
the maxima and minima of a function subject
to constraints.

3.  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.

4. Lets consider there is a function f(x, y, z)whose maxima or
minima is to be find out subjected to the constraint
g(x, y, z) ..
 Solve the following system of equations.
f(x, y, z) = λ g(x, y, z)
g(x, y, z) = k
• 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 constant, , is called the Lagrange Multiplier.

6.  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

7. • 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)

8.  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)

9.  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

10.  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.

11.  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.

12. Here,
f(x1,x2,x3,x4,x5)= x1
2+x2
2+x3
2+x4
2+x5
2
Constraints
g1(x1,x2,x3,x4,x5)=x1+2x2+x3-1
g2(x1,x2,x3,x4,x5)=x3-2x4+x5-6
Thus,
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

13. δg1(x1,x2,x3,x4,x5)/δ x1 =1
δg1 (x1,x2,x3,x4,x5)/δx2 =2
δg1 (x1,x2,x3,x4,x5)/δx3 =1
δg1 (x1,x2,x3,x4,x5)/δx4 =0
δg1 (x1,x2,x3,x4,x5)/δx5 =0
δg2(x1,x2,x3,x4,x5)/δ x1 =0
δg2 (x1,x2,x3,x4,x5)/δx2 =0
δg2 (x1,x2,x3,x4,x5)/δx3 =1
δg2 (x1,x2,x3,x4,x5)/δx4 =-2
δg2 (x1,x2,x3,x4,x5)/δx5 =1
On 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

14. 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 x5
We 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)

15. On Solving these equations
µ=-2, λ=0
Plugging these into the equations we get
x1=x2=0, x3=x5=1 and x4=-2
To 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.

16. Here,
f(x , y , z)=4y-2z
Constraint,
g1(x ,y ,z)=2x-y-z-2
g2(x ,y ,z)= x² + y² -1
Thus,
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

17. δg1(x ,y ,z)/δz =-1
δg2(x ,y ,z)/δx=2x
δg2(x ,y ,z)/δy =2y
δg2(x ,y ,z)/δz=0
On 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)/δz
We 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)

18. On Solving these equations…
λ=2, µ=+5,-5
Plugging these into the equations we get
If µ=+5 ,then x=0.8,y=-0.6,z=0.2 and
If µ=-5 ,then x=-0.8,y=0.6,z=-4.2
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(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)