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1)Domain for optimization

1)Domain for optimization. Done by: Fatema Al Hebabi Student ID: 200802575 Serial no:05. GOAL: Determine your goal; whether you want to maximize of minimize. DATA : Introduce variables and given values. EQUATIONS : introduce the necessary adequate equations.

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1)Domain for optimization

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  1. 1)Domain for optimization Done by: Fatema Al Hebabi Student ID: 200802575 Serial no:05

  2. 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. Steps of Optimization

  3. ** 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.

  4. 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 .

  5. 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 .

  6. 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

  7. So Critical points are x=25 or x=8.33 P’’(x)= 6x-100 • Note that since both x and y are nonnegative numbers and their sum is 25, it follows that . See the adjoining sign chart for P' .

  8. + 0 - P’ X=25 Y=0 P=0 X=0 Y=25 P=0 X=8.33 Y=25 P=1225 • If • x=25 and y=8.33 , • then • P= 1225 • is the largest possible product.

  9. P(x) P’(x) P’’(x)

  10. By: Fatema Mesfer Al-hebabi200802575

  11. 2)Optimization Steps OMAIMA DAMMAK QUID : 201000904 Serial no: 25

  12. DETERMINE YOU GOAL OPTION 1 : To maximize OPTION 2: To minimize GOAL

  13. Introduce the given variables and values. DATA

  14. Introduce the adequate equations . EQUATIONS

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

  16. Differentiate the combined equation to the 1st and 2nd derivatives. • F’(x) • F’’(x) DIFFERENTIATE

  17. Use 1st and 2nd derivatives to: • Determine critical points. (check end points if applicable) • Determine whether a local min and max exist . EXTREMA

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

  19. 3)Optimization problems Nourhankhalilabdo Student ID:200907407 Serial no: 01

  20. 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? Example

  21. River X X 500-2X

  22. 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 Solution

  23. 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

  24. 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

  25. MAXIMUM VALUES ++++++++++++++++++++ --------------------------------- 0 125 300 X=1 X=200

  26. 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 ………

  27. 4)Worked examples on optimization Hiba Abu Watfa 200911397 Serial no: 03

  28. problem number (5) From Section 4.7 in the book:- Find the dimensions of a rectangle with perimeter 100 m whose area is as large as possible.

  29. Solving the Problem Apply Convert Understand • 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

  30. Problem illustration:- Area 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. Maximization

  31. Solution:- Rectangle perimeter =2(x+y)* 100=2x+2y Y=(100-2x)/2 Y=50-x (1) *Rectangle Area =xy A=xy (2)

  32. 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).

  33. 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.

  34. 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.

  35. The clue Finally, the rectangle dimensions are 25m and 25 m and the maximum area possiple is 625 m2.

  36. Sketching the graph

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