Math notes Essay

Unit #23 :

Lagrange Multipliers

Goals:
• To study constrained optimization; that is, the maximizing or minimizing of a function subject to a constraint (or side condition).
Reading: Section 15.3.

Constrained Optimization - Examples - 1

In the previous section, we saw some of the diﬃculties of working with optimization when there are multiple variables. Many of those problems can be cast into an important class of problems called constrained optimization problems, which can be solved in an alternative way.
Examples of Problems with Constraints
1. Show that the rectangle of maximum area that has a given perimeter p is a square.
The function to be maximized:
A(x, y) = xy
The constraint:
2x + 2y = p

Constrained Optimization - Examples - 2

2. Find the point on the sphere x2 + y 2 + z 2 = 4 that is closest to the point
(3, 3, 5).
Here we want to minimize the distance
D(x, y, z) =

(x − 3)2 + (y − 3)2 + (z − 5)2

subject to the constraint x2 + y 2 + z 2 = 4

Lagrange Multipliers - 1

Lagrange Multipliers
To solve optimization problems when we have constraints on our choice of x and y, we can use the method of Lagrange multipliers. Suppose we want to maximize the function f (x, y) subject to the constraint g(x, y) = k. Consider the relative positions of their level curves:

Lagrange Multipliers - 2

Lagrange Multipliers - 3

On the above diagram, locate the maximum and minimum of f (x, y) (label with M and m respectively) on the level curve g(x, y) = k.
Draw in the gradient vectors ∇f and ∇g at M and m. Describe, in words, the relationship between ∇f (M ) and ∇g(M ), and also between ∇f (m) and
∇g(m).

The Lagrange Multiplier Method - 1

Description of the Lagrange Multiplier Method
Step 1. Find all values of (x, y) and λ such that
∇f (x, y) = λ∇g(x, y) g(x, y) = k
Step 2. Evaluate f at all the points (x, y) obtained in Step 1. (The largest value is the maximum and the smallest is the minimum.)

The Lagrange Multiplier Method - 2

Note that Step 1 really amounts to solving three equations in three unknowns (x, y, and λ). The equations can be found by rewriting the gradient equations as follows: fx(x, y) = λgx(x, y) fy (x, y) = λgy (x, y) g(x, y) = k
The same method applies to functions of three (or more) variables. In this case of three variables, we would solve four equations in four unknowns.

Lagrange Multiplier Method - Linear Constraint - 1

Example: values of

Consider the problem of ﬁnding the maximum and minimum

x2 + y 2
,
f (x, y) = x − 10y +
20
subject to the constraint x + y = 10 .
Sketch the meaning of the constraint.

Lagrange Multiplier Method - Linear Constraint - 2

Write down the three equations obtained by the method of Lagrange multipliers.

Lagrange Multiplier Method - Linear Constraint - 3

Solve these equations, and compare the values at the resulting points to ﬁnd the maximum and minimum values.

Lagrange Multiplier Method - Linear Constraint - 4

70

80

−3

64
50

−3

62

30

60

40

60

−3

66
60

50 5
−3 −34 340 35
− −3

30

−3

68
70

55
−3
−3 50
−3 45
−3 40
35

58
30

−3

30

56
20

54
10

0
−90

52

−80

−70

−60

−50

−40

−30

−20

−10

0

10

50

−60

−55

−50

−3
−3 60
55
−3
−3 50
−3 45
−3 40
35

−3
−3 60
55
−3
−3 50
−3 45
−3
30 − 40
33
5
−45

−40

Lagrange Multiplier Method - Non-Linear Constraint - 1

Example: Find the points on the curve x4 + y 4 = 1 that are closest to and furthest from the origin.

Lagrange Multiplier Method - Non-Linear Constraint - 2

x4 + y 4 = 1

Lagrange Multiplier Method - Non-Linear Constraint - 3

x4 + y 4 = 1

Lagrange Multiplier Method - Non-Linear Constraint - 4

1.5

1.5

1

1

0.5

0.5

0

0

−0.5

−0.5

−1

−1

−1.5

−1.5

−1

−0.5

0

0.5

1

1.5

−1.5

−1

−0.5

−1.5

−1.5

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