mth374: optimization for master of mathematics by dr. m. fazeel anwar assistant professor department...
Post on 24-Dec-2015
223 Views
Preview:
TRANSCRIPT
1
MTH374: Optimization
For
Master of Mathematics
By
Dr. M. Fazeel AnwarAssistant Professor
Department of Mathematics, CIIT Islamabad
2
Lecture 04
3
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
15
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?
27
28
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
30
31
Summary
Thank You
top related