by Dr. Shreyasi Jana

Unit I(C9T, SEM-IV), Maxima and Minima of Several Variables,Lagrange Multiplier and Constrained Optimization

byDr. Shreyasi Jana

Asst. Prof., Department of MathematicsNarajole Raj College

WB, India

Maxima and Minima for several variablesFinding relative extrema

ExamplesLagrange Multiplier and Constrained Optimization Problem

ProblemsExercise

Outline

1 Maxima and Minima for several variables

2 Finding relative extrema

3 Examples

4 Lagrange Multiplier and Constrained Optimization Problem

5 Problems

6 Exercise

Unit I(C9T, SEM-IV), Maxima and Minima of Several Variables, Lagrange Multiplier and Constrained Optimization by Dr. Shreyasi Jana



ProblemsExercise

Maxima and Minima for one variable

For a function of one variable, f(x), we find the local maxima/minima by differentiation.Maxima/minima occur when f ′(x) = 0.

Maxima and Minima

x = a is a maximum if f ′(a) = 0 and f ′′(a) < 0.

x = a is a minimum if f ′(a) = 0 and f ′′(a) > 0.

Note: A point where f ′′(a) = 0 and f ′′′(a) 6= 0 is called a point of inflection.Geometrically, the equation y = f(x) represents a curve in the two-dimensional (x, y)plane, and we call this curve the graph of the function f(x).




ProblemsExercise

Maxima and Minima for two variables

Definition

A function f(x,y) of two variables is said to have a maximum (minimum) at a point (a,b) if there is a disc centered at (a, b) such that

f (a, b) ≥ f (x , y), (f (a, b) ≤ f (x , y))

for all points (x, y) that lie inside the disc.

The extreme-value theorem

If f(x, y) is continuous on a closed and bounded set R, then f has both maximum andminimum on R.




ProblemsExercise

Local and global maxima and minima

Local and global maxima and minima for cos(3πx)/x , 0.1 ≤ x ≤ 1.1.




ProblemsExercise

Necessary condition for local Maxima and Minima

The extreme-value theorem

If f(x,y) has a local extremum at (a, b), and if the first-order derivatives of f exist atthis point, then

fx (a, b) = 0 and fy (a, b) = 0.

Critical Points

A point (a, b) in the domain of f(x, y) is called a critical point of f if fx (a, b) = 0 andfy (a, b) = 0, or if one or both partial derivatives do not exist at (a, b).




ProblemsExercise

Saddle Point

Saddle Point

A saddle point or minimax point is a point on the surface of the graph of a functionwhere the slopes (derivatives) in orthogonal directions are all zero (a critical point),but which is not a local extremum of the function. An example of a saddle point iswhen there is a critical point with a relative minimum along one axial direction(between peaks) and at a relative maximum along the crossing axis.

Example

The function f (x , y) = x2 + y3 has a critical point at (0,0) that is a saddle point sinceit is neither a maximum nor minimum.




ProblemsExercise

Saddle Point on the graph z = x2 − y2

The red point indicates the saddle Point on the graph z = x2 − y2.




ProblemsExercise

Saddle Point on the graph y = x3

The plot of y = x3 with a saddle point at 0.




ProblemsExercise

Second Partial Derivative Test

Second Partial Derivative Test

Let f(x, y) have continuous second-order partial derivatives in some disc centred at acritical point (a, b), and let D = fxx (a, b)fyy (a, b)− (fxy (a, b))2

If D > 0 and fxx (a, b) > 0, then f has a relative minimum at (a, b).

If D > 0 and fxx (a, b) < 0, then f has a relative maximum at (a, b).

If D < 0, then f has a saddle point at (a, b).

If D = 0, then no conclusion can be drawn.




ProblemsExercise

Example 1

Example

1. Find and classify all the critical points for 3x2y + y3 − 3x2 − 3y2 + 2.

Here fx = 6xy − 6x , fy = 3x2 + 3y2 − 6y , fxx = 6y − 6, fyy = 6y − 6, fxy = 6x . Forcritical points we solve 6xy − 6x = 0 and 3x2 + 3y2 − 6y = 0.Solving we get the critical points are (0,0), (0,2), (1,1), (-1,1). Now

D = fxx (a, b)fyy (a, b)− (fxy (a, b))2 = (6y − 6)2 − 36x2.

at (0,0), D = 36 > 0, fxx (0, 0) = −6 < 0, Relative Maximum.

at (0,2), D = 36 > 0, fxx (0, 2) = 6 > 0, Relative Minimum.

at (1,1), D = −36 < 0, Saddle Point.

at (-1,1), D = −36 < 0, Saddle Point.




ProblemsExercise

Example 2

Example

2. Determine the point on the plane 4x − 2y + z = 1 that is closest to the point(-2,-1,5).

Note that we are NOT asking for the critical points of the plane. First, lets supposethat (x,y,z) is any point on the plane. The distance between this point and the pointin question, (-2,-1,5), is given by the formula,

d =√

(x + 2)2 + (y + 1)2 + (z − 5)2.

What we are then asked to find is the minimum value of this equation. The point(x,y,z) that gives the minimum value of this equation will be the point on the planethat is closest to (-2,-1,5).Now it is a function of x, y ,z and we can only deal with functions of x and y at thispoint. We can solve the equation of the plane to see that, z = 1− 4x + 2y .




ProblemsExercise

Example 2 Continued

Example

Plugging this into the distance formula gives,d =

√(x + 2)2 + (y + 1)2 + (1− 4x + 2y − 5)2

=√

(x + 2)2 + (y + 1)2 + (−4− 4x + 2y)2.Lets notice that finding the minimum value of d will be equivalent to finding theminimum value of d2. So, lets instead find the minimum value of

d2 = (x + 2)2 + (y + 1)2 + (−4− 4x + 2y)2.

Here fx = 36 + 34x − 16y , fy = −14− 16x + 10y , fxx = 34, fyy = 10, fxy = −16.Now

D = fxx (a, b)fyy (a, b)− (fxy (a, b))2 = 84 > 0.




ProblemsExercise

Example 2 Continued

Example

So D > 0, also fxx > 0 and so any critical points that we get will be guaranteed to berelative minimums.For critical points we solve

36 + 34x − 16y = 0

and−14− 16x + 10y = 0.

Solving we get the critical point is (− 3421,− 25

21).

We can find the z coordinate by plugging into the equation of the planez = 1− 4x + 2y = 107

21.

So, the point on the plane that is closest to (-2,-1,5) is (− 3421,− 25

21, 107

21).




ProblemsExercise

Lagrange Multiplier

The method of Lagrange multipliers is a strategy for finding the local maxima andminima of a function subject to equality constraints. Consider the optimization problem

max f (x , y)subject to g(x , y) = 0.We assume that both f and g have continuous first partial derivatives. We introducea new variable λ called a Lagrange multiplier and study the Lagrange function (orLagrangian) defined by L(x , y , λ) = f (x , y) − λg(x , y). If f (x0, y0) is a maximum off(x,y) for the original constrained problem, then there exists λ0 such that (x0, y0, λ0) isa stationary point for the Lagrange function L.

Lagrange Multiplier Method

Suppose f and g have continuous partial derivatives. Let(x0, y0) ∈ S = {(x , y) : g(x , y) = 0} and ∇g(x0, y0) = 0. If f has a local maximum orminimum at (x0, y0) then there exists λ ∈ R such that

∇f (x0, y0)− λ∇g(x0, y0) = 0.




ProblemsExercise

Lagrange Multiplier

The red curve shows the constraint g(x , y) = c. The blue curves are contours of f(x, y).The point where the red constraint tangentially touches a blue contour is the maximumof f(x, y) along the constraint, since d1 > d2.




ProblemsExercise

Problem 1

Example

1. Find the maximum and minimum of f (x , y) = 5x − 3y subject to theconstraint g(x , y) = x2 + y2 = 136.

Here the system that we need to solve is

∇f (x0, y0)− λ∇g(x0, y0) = 0.

This gives5 = 2λx ,

−3 = 2λy ,

x2 + y2 = 136.

Solving this we get λ = ± 14.




ProblemsExercise

Problem 1 Continued

Example

If λ = − 14

, we get x = −10, y = 6.

If λ = 14

, we get x = 10, y = −6.

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).




ProblemsExercise

Exercise

Find and classify all the critical points of the following functions.

1. f (x , y) = 2y − 9x − xy + 5x2 + y2

2. f (x , y) = x3 − y3 + 8xy

3. f (x , y) = (y − x)(1− 2x − 3y)

4. f (x , y) = xye−8(x2+y2).

Lagrange Multiplier

1. Find the maximum and minimum of f (x , y) = 4x2 + 10y2 on the diskx2 + y2 ≤ 4.

2. Find the maximum and minimum of f (x , y , z) = 4y − 2z subject to theconstraints 2x − y − z = 2, x2 + y2 = 1.

3. Find the maximum and minimum of f (x , y , z) = xyz subject to the constraintx2 + 2y2 + 4z2 = 24.

4. Determine the point on the plane 2x + 3y − z = 5 that is closest to origin.


Thank You

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by Dr. Shreyasi Jana