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MTH374: Optimization

For

 Master of Mathematics

By

Dr. M. Fazeel AnwarAssistant Professor

Department of Mathematics, CIIT Islamabad

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Lecture 04

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Recap

• Optimization problem• Variables and objective functions• Some optimization from calculus

A general optimization problem

An optimization problem can be stated as follows:

Find which minimizes a function

subject to the constraints

for

for

for .

Some notations

• The variable is called a design vector or decision variable.

• The function is called the objective function.• The functions are called the constraints of the problem.• The problem is called a constrained optimization

problem.• If there are no constraints then the problem is called an

unconstrained optimization problem.

Today’s Topics

• Some optimization from calculus• One variable optimization• Multivariable optimization

Relative Maxima and Minima

1. A function is said to have a relative maximum at if there is an open interval containing on which is the largest value, that is, for all in the interval.

2. A function is said to have a relative minimum at if there is an open interval containing on which is the smallest value, that is, for all in the interval

Note: If has a relative maximum or a relative minimumat , then is said to have a relative extremum at

Relative Maxima and Minima

Example

Critical PointsA critical point for a function is a point in the domain of at which either the graph of has a horizontal tangent line or is not differentiable.

A critical point is called a stationary point of if

Relative extrema and critical pointsSuppose that is a function defined on an open interval containing the point If has a relative extremum at then is a critical point of i.e. either or is not differentiable at

Example

xxxf 123)( 2

0123 2 xx

Find all the relative extrema of

0)4(3 xx04or 03 xx

4,0x

Relative max. Relative min.

Critical points

(0) 1f 3 2(4) (4) 6(4) 1 31f

First Derivative TestSuppose that f is continuous at a critical point 0.x

1. If ( ) 0f x on an open interval extending left from 0x and

( ) 0f x on an open interval extending right from 0 ,x then

f has a relative maximum at 0.x

2. If on an open interval extending left from ( ) 0f x 0x

0.x0 ,x

and

( ) 0f x on an open interval extending right from thenf has a relative minimum at

3. If ( )f x has same sign on an open interval extending left

from 0x as it does on an open interval extending right from 0 ,x then f does not have a relative extrema 0.x

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Use the first derivative test to show that Example

2( ) 3 6 1f x x x has a relative minimum at x=1

2( ) 3 6 1f x x x ( ) 6 6f x x

6( 1)x

f has relative minima at x=1

Solution

x=1 is a critical point as ( ) 0 at 1.f x x

Interval

Test Value

Sing of - +

Conclusionis decreasing on 

is decreasing on 

1x 1x

c 0 2

( )f c

(0) 6 0f (2) 6 0f ( )f c

f1x

f1x

Second Derivative TestSuppose that f is twice differentiable at the 0.x

(a) If 0( ) 0f x and 0( ) 0,f x then f has relative minimum at 0.x

(b) If 0( ) 0f x and 0( ) 0,f x then f has relative minimum at 0.x

(c) If 0( ) 0f x and 0( ) 0,f x then the test is inconclusive;

that is, f may have a relative maximum, a relative minimum,

or neither at 0.x

ExampleFind the relative extrema of 5 3( ) 3 5f x x x

Solution4 2( ) 15 15f x x x

2 215 ( 1)x x

2( ) 15 ( 1)( 1)f x x x x

3( ) 60 30f x x x 230 (2 1)x x

Critical Points

Setting ( ) 0f x

215 ( 1)( 1) 0x x x 215 0 or 1 0 or 1 0x x x

1,0,1x are critical points

Stationary Point

Second Derivative Test

-30 -has a relative

maximum

0 0 Inconclusive

30 + has a relative

minimum

230 (2 1)x x ( )f x

1x

f

0x

1x

f

Absolute Extrema

Let I be an interval in the domain of a function f. We say that f has an absolute maximum at a point in I if

for all x in I, and we say that f has an absolute minimum at if for all x in I.

0x0x 0( ) ( )f x f x

0( ) ( )f x f x

Extreme value Theorem

If a function f is continuous on a finite closed interval [a, b] then f has both an absolute maximum and an absolute minimum on [a, b].

Procedure for finding the absolute extrema of a continuous function f on a finite closed interval [a, b]

Step 1. Find the critical points of f in (a, b).

Step 2. Evaluate f at all the critical points and at the end points a and b.

Step 3. The largest of the value in step 2 is the absolute maximum value of f on [a, b] and the smallest value is the absolute minimum

Find the absolute maximum and minimum values of the function

Example

3 2( ) 2 15 36f x x x x on the interval [1, 5], and determine where these values occur.

solution2( ) 6 30 36f x x x

2( ) 6( 5 6)f x x x

6( 2)( 3)x x

( ) 0f x at x=2 and x=3

So x=2 and x=3 are stationary points

Evaluating f at the end points, at x=2 and at x=3 and at the endspoints of the interval.

3 2(1) 2(1) 15(1) 36(1) 23f

3 2(2) 2(2) 15(2) 36(2) 28f

3 2(3) 2(3) 15(3) 36(3) 27f

3 2(5) 2(5) 15(5) 36(5) 55f

Absolute minimum is 23 at x=1

Absolute minimum is 55 at x=5

Absolute extrema on infinite intervals

Absolute extrema on infinite intervals

Example (Solution)

• A rectangular plot of land is to be fenced in using two kinds of fencing. Two opposite sides will use heavy duty fencing selling for $3 a foot, while the remaining two sides will use standard fencing selling for $2 a foot. What are the dimensions of the rectangular plot of greatest area that can be fenced in at a cost of $6000?

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Example. (Solution)

• Suppose that the number of bacteria in a culture at time is given by Find the largest and smallest number of bacteria in a culture during the time interval

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Summary

Thank You