Optimization

Optimization (mathematics)

In mathematics, the simplest case of optimization, or mathematical

programming, refers to the study of problems in which one seeks to minimize or maximize a real function by systematically

choosing the values of real or integer variables from within an allowed set.

In mathematics, maxima and minima, known collectively as extrema, are the largest value (maximum) or smallest value (minimum), that a function takes in a point either within a given neighbourhood (local extremum) or on the function domain in its entirety (global extremum).

We frequently encounter situations in which we are asked to do the best we can. Such a request is vague unless we are given some conditions.

• Asking us to minimize the cost of making tables and chairs is not clear.

• Asking us to make the maximum number of tables and chairs possible so that the costs of production are minimized and given that the amount of material available is restricted allows us to construct a function of the situation. We can then determine the minimum or maximum of the function

• Such a procedure is called optimization

• To optimize a situation is to realize the best possible outcome, subject to a set of restrictions.

• Because of these restrictions, the domain of the function is usually restricted.

• The max or min value can be determine through calculus, or might also occur at the ends of the domain.

and

Problems

(p,q)

Remember the graph of y = x2

(a basic quadratic)examine the nature of the

graphs(p,q)

If we could model a “real life situation” using a function, we could determine a maximum or minimum value for that situation ….

Farmer Brown has hit hard times …

He and his dog Bingo can only afford 80m of fencing.

He needs to maximize the area of land his cattle can feed upon.

What should the dimensions of the lot be?

5m 5m

70m

A = 5m X 70m = 350m2

P = 5m + 70m + 5m = 80m

8m 8m

64 m

A = 8m X 64m = 512m2

P = 8m + 64m + 8m = 80m

12m 12m

56 m

A = 12m X 56m = 672m2

P = 12m + 56m + 12m = 80m

Let’s try to model this situation:

1. Maximize the Area : A

2. Let x = width, let y = length

A = (w)(l)

A = (x)(y)

Xm Xm

Ym

Let’s try to model this situation:

3. The Link (second equation):

P = 80 = 2w + l

= 2x + y = 80

Using the 2 equations and substitution, we can create a quadratic equation

y = 80 – 2x

A = (x)(y) 2x + y = 80

A = (x)(80 – 2x)

= 80x – 2x2

= -2x2 + 80x

Complete the square to convert, or…….

A = -2x2 + 80x

Find the change in Area with respect to the change in x.

dx

xxd

dx

dA )802( 2

804 xMake the derivative equal to zero and solve for x (to determine the max point)

8040 x

Find y

20x

2x + y = 80

2(20) + y = 80

y = 40

The width

The length

20m 20m

40 m

A = 20m X 40m = 800m2

P = 12m + 56m + 12m = 80m

Consider a dance recital…

At $3.00 a ticket, all 800 seats will be sold.

For every $1.00 increase in ticket price, 100 fewer seats will be sold.

What ticket price will maximize the profit?

Try a few calculations:

R = (3)(800) = $2400.00

Increase by $1.00

R = (4)(700) = $2800.00

Increase by $7.00

R = (10)(100) = $1000.00

1. Maximize the Revenue : R

2. Let x be the number of dollar increases in the ticket price.

R = (cost per ticket)(number of tickets sold)

= (3)(800)

= (3 + x)(800 – 100x)

= -100x2 + 500x + 2400

Find the change in revenue with respect to the increase in ticket price.

dx

xxd

dx

dR )2400500100( 2

500200 xdx

dR Equate to zero and solve for x

0 = -200x + 5002.5 = x Sub to determine the

maximum Revenue$3025 = R

3. The equation already has 1 variable

4. Complete the square to convert

R = -100x2 + 500x + 2400

= -100(x2 – 5x) + 2400

= -100(x2 – 5x +6.25 – 6.25) + 2400

= -100(x2 – 5x + 6.25) + 625 + 2400

= -100(x – 2.5)2 + 3025

sheet

1,2,4,9,1013,15

Pg 204 read example 1

Pg 205 read example 2

Pg 206

1,4,6,7,9,11