1)domain for optimization

1)Domain for optimization

Done by: Fatema Al HebabiStudent ID: 200802575

Serial no:05

Steps of Optimization

GOAL: Determine your goal; whether you want to maximize of minimize.

DATA : Introduce variables and given values.EQUATIONS : introduce the necessary adequate equations.COMBINE : combine them to obtain a differentiable , single

variable equation.DIFFERENTIATE : Use your calculus skills to differentiate

the obtained equation into first and then second derivative.EXTREMA: Use 1st and 2nd derivatives to:

**Determine critical points. (check end points if applicable)**Determine whether a local min and max exist .

CONCLUSION : substitute the obtained ''x'' value in the COMBINED equation to obtain your goal . Afterwards you can easily get the other unknown values.

**Find two nonnegative numbers whose sum is 25 and so that the product of one number and the square of the other

number is a maximum .

SOLUTION : Let variables x and y represent two

nonnegative numbers. The sum of the two numbers is given to be

25 = x + y , so that

y = 25 - x . We wish to MAXIMIZE the PRODUCT

P = x y2 .

However, before we differentiate the right-hand side, we will write it as a function of x only. Substitute for y getting

P = x y2 = x (25 -x)2 .

Now differentiate this equation using the product rule and chain rule, getting

P‘= x (2) ( 25-x)(-1) + (1) ( 25-x) 2

= 1.(25-x) 2 +x.(-2 (25-x)) = (25-x) 2 -2x.(25-x)= 3x 2 - 100x +625

So Critical points are x=25 or x=8.33

P’’(x)= 6x-100Note that since both x and y are nonnegative

numbers and their sum is 25, it follows that . See the adjoining sign chart for P' .

If x=25 and y=8.33 , then P= 1225is the largest possible product.

P’ +0 -

X=0Y=25P=0

X=8.33Y=25P=1225

X=25Y=0P=0

P(x) P’(x) P’’(x)

By: Fatema Mesfer Al-hebabi200802575

2)Optimization StepsOMAIMA DAMMAKQUID : 201000904

Serial no: 25

GOAL

DETERMINE YOU GOAL

OPTION 1 : To maximize

OPTION 2: To minimize

DATA

Introduce the given variables and values.

EQUATIONS

Introduce the adequate equations .

COMINATION

Combine those equations to obtain a differentiable , single valued equation.

DIFFERENTIATE

Differentiate the combined equation to the 1st and 2nd derivatives.

F’(x)F’’(x)

EXTREMA

Use 1st and 2nd derivatives to:◦Determine critical points. (check end points if

applicable)◦Determine whether a local min and max exist .

CONCLUSION

Substitute the obtained ''x'' value in the COMBINED equation to obtain your goal . Afterwards you can easily get the other unknown values.

3)Optimization problems

Nourhan khalil abdoStudent ID:200907407

Serial no: 01

Example

A man has a farm that is adjacent to a river. Suppose he want to build a rectangle pen for his cows with 500 ft of fencing.

If one side of the pen is river ((his cows will not swim away???)),

What is the area of the largest pen he can build?

River

XX

500-2X

Solution

Firstly , man need to build pen for his farm to protect his cows 500-2x

As we see in one side of his farm is river.. So he need to calculate the maximum area of the largest pen he should be build ?

So let we say one side of this rectangle is X. So in front of this side should be also X, because it is a

rectangle and in rectangle there are 4 sides and every two sides are similar to each other

As they gave us, they say the parameter is equal to 500 ft So if two sides of rectangle are X That mean X+X =2X And then other side equal 500-2X

As we know we the area of any rectangle equal = length x width

So A(x) = X (500-2X) …………………………………. (1) Which equal A(x) = 500X-2X2 …………… (1) After that we should find the derivatives of our area to can

find the maximum area of the largest pen??? A’(x) = 500-4X ………………………………..(2) As my partners explain we can find the maximum or

minimum values from critical numbers So, We can get the critical numbers when A’(x) = 0 or A’(x) = undefined

When A’(x) = 0 500-4X =0 Then make all variables x in one side 4X = 500 sooo X = 500/4 X = 125 From this point we can make the test line which makes more clear

the maximum values>> When X =1 …………………………….(2) A’ n(1)= 500 -4(1) = 496 ………………………. Positive point When x= 200 ((we just choose any point not more than 500))

…………………………….(2) A’ (200) = 500-4(200) = -300 ………………………..negative point

0 125 300++++++++++++++++++

++---------------------------------

MAXIMUM VALUES

X=1 X=200

after we get the test line in which point is increasing or decreasing

We just substitute that point which we do the test about them which equal 125

We put this point in our equations …. (1)A(x) = X (500-2X)A (125) =125(500-2*125) =31,250 ft2

By this way we determined what is the area he need it to build the largest pen ………

4)Worked examples on optimization

Hiba Abu Watfa200911397

Serial no: 03

Find the dimensions of a rectangle with perimeter 100 m whose area is as large as possible.

problem number (5)From Section 4.7 in the book-:

Solving the Problem

ConvertApplyUnderstand

Understand the problem Convert the (Physics, economics, etc)

problems to mathematic problems Apply the technique of the preceding section to obtain the max or min

Problem illustration-:

What we wish to do in here is to get the maximum possible area A of a rectangle with a 100 m perimeter in order to get the unknown dimensions of it .

Area

Maximization

Solution-:Rectangle perimeter =2(x+y) *

100=2x+2yY=(100-2x)/2Y=50-x (1)

*Rectangle Area =xyA=xy (2)

Continuing the solution

*Expressing Y in terms of X: from 1 & 2:A(x)= x(50-x)A= 50x –x2 (3) *The function which we wish to maximize

* The domain of the function is 0 ≤ X ≤ 25 (Otherwise A < 0)*The derivative is A'(x) = 50 -2x = 2(25-x).

continue

*Solving the equation in order to find the critical numbers which gives x = 25.Because the maximum value of A must occur either at this critical number or at an end point of the interval, and since A(0)=0, A(25)=625, and A(50)=0, the closed interval method gives the maximum value as A(25)=625.

continue*OR we could find the second

derivative which is A"(x) = -2 which is < 0 for all x, and from that we can conclude that A is always concave downward and the local maximum at x =25 must be an absolute maximum.

Finally, the rectangle dimensions are 25m and 25

m

The clue

and the maximum area possiple is 625 m2 .

Sketching the graph

0 10 20 30 40 50 600

100200300400500600700

A(x) =50x -x2

X

Y