This document discusses the Lagrange multiplier method for finding the constrained maximum or minimum of a function subject to an equality constraint. It provides examples of using Lagrange multipliers to find the dimensions of a rectangle with maximum area given a perimeter, and to find the points on a circle closest to and farthest from a given point. The key steps are to set up the Lagrange multiplier equation relating the gradients of the objective function and constraint, solve for the critical points, and evaluate the objective function at these points to find the maximum or minimum.

1. Lagrange Multiplier
Dr. Anjali Devi J S
Guest Faculty
School of Chemical Sciences
M G University
1

2. Opimisation of a function
Find maxima and minima of the function
To find maxima/minima of a function f(x):
d[f(x)]/dx= f’(x) =0
at x=a
f’’(a)>0 , a is minimum point
f’’(a)< 0 , a maximum point
2

3. Identify the absolute extrema and relative extrema for the following
function.
f(x)=x2 on[−1,2]
Question
Relative and absolute minimum of
zero at x=0 and an absolute
maximum of four at x=2
Answer
3

4. Constrained Opimisation of functions
Find maxima and minima of the function with constraints on the
variable
For a rectangle whose perimeter is 20 m, find the dimensions that will
maximize the area.
Question
breadth= x
length= y
Area, A =xy
Perimeter,P =2x+2y
Maximise: f(x,y) = xy
Given: 2x+2y =20
4

5. Find maxima and minima of the function with constraints on the
variable
For a rectangle whose perimeter is 20 m, find the dimensions that will
maximize the area.
Question
Constrained Opimisation of functions
f (x,y) =xy (1)
2x+2y =20 (2)
From (2), y=10-x
Substituting y,
f(x,y) = 10x-x2
On the interval [0,10]
f’(x) =10-2x=0
x=5
f’’(5)=-2<0
breadth= x=5
length= y=5
X=y=5 is a
maximum point
5

6. Constrained Opimisation
Maximise (or minimise): f(x,y) (or f(x,y,z))
Given: g(x,y)=c
(Or g(x,y,z)=c) for some constant
g(x,y) =c is constraint equation.
Or we say that x and y are constrained by g(x,y)=c.
Points (x,y) which are maxima or minima of f(x,y) with the condition
that they satisfy the constraint equation g(x,y)=c are
called constrained maximum or constrained minimum points,
respectively. Similar definitions hold for functions of three variables.6

7. Lagrange Multipliers method
Let f(x,y) and g(x,y) be smooth functions, and suppose that c is a scalar
constant such that ∇g(x,y)≠0 for all (x,y) that satisfy the
equation g(x,y)=c. Then to solve the constrained optimization problem
Maximize (or minimize) : f(x,y)
given : g(x,y)=c,
find the points (x,y) that solve the equation ∇f(x,y)=λ∇g(x,y) for some
constant λ (the number λ is called the Lagrange multiplier). If there is a
constrained maximum or minimum, then it must be such a point.
7

8. For a rectangle whose perimeter is 20 m, find the dimensions that will
maximize the area.
Question
length= y
breadth= x
Area ,A=xy
Perimeter,P =2x+2y
Maximise: f(x,y) = xy
Given: 2x+2y =20
∇f(x,y)=λ∇g(x,y)
∂f/∂x=λ∂g/∂x and ∂f/∂y=λ∂g/∂y
y=2λ,
x=2λ i.e., x=y
Substitute either of the
expressions for x or y into
the constraint equation
20=g(x,y)=2x+2y=2x+2x=4x
⇒x=5
⇒y=5
∴ The maximum area occurs for a rectangle whose width and height both
are 5 m. 8

9. Maximixe (Minimise): f(x,y) =(𝑥 − 1)2+(𝑦 − 2)2
Given: g(x,y) = 𝑥2 +𝑦2= 80
Answer
Distance, d, between points (x,y) and
(1,2) == (𝑥 − 1)2+(𝑦 − 2)2
∇f(x,y)=λ∇g(x,y)
∂f/∂x=λ∂g/∂x and ∂f/∂y=λ∂g/∂y
2 (x-1) =2λx,
2 (y-2)=2λy
i.e., y=2x
Substitute this into
g(x,y) = 𝑥2
+𝑦2
= 80
5𝑥2= 80
𝑥 = ∓4
x
y
(4,8)
(1,2)
(-4,-8)
f(4,8)=45 ,
f(-4,-8) = 125 9
Question
Find the points on the circle x2+y2=80 which are closest to and farthest
from the point (1,2).

10. 10
Question (Continued)
Find the points on the circle x2+y2=80 which are closest to and farthest
from the point (1,2).
∴ The two critical points are
(4,8) and (-4,-8).
f(4.8)< f(-4,-8)
Constrained Maximum point: (-4,-8)
Constrained Minimum point: (4,8)

11. Maximize (and minimize) : f(x,y,z)=x+z
given :g(x,y,z)=x2+y2+z2=1
Answer
Question
∇f(x,y)=λ∇g(x,y)
∂f/∂x=λ∂g/∂x and ∂f/∂y=λ∂g/∂y
1=2λx
0=2λy
1=2λz
y=0
x=
1
2λ
=z
Substitute this into constraint
equation g(x,y,z)=x2+y2+z2=1
Critical points: (
1
2
, 0,
1
2
)
and (
−1
2
, 0,
−1
2
)
f (
1
2
, 0,
1
2
) > f (
−1
2
, 0,
−1
2
)
Constrained Maximum point: (
1
2
, 0,
1
2
)
Constrained Minimum point : (
−1
2
, 0,
−1
2
)
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