Add Math Project 2015Title: Linear ProgrammingStudent: Chin Pei FooClass: 5N

Content Index Page

Title1

Content 2

Pierre de Fermat3

Definition 4

Method of quadratic equation5

Sheep farming8

Box9

Rush Hours10

Linear Programming12

Daily life example of LP14

Further exploration16

Graph17

Cabinet18

Conclusion21

Pierre de FermatPierre de Fermat (French:[pj dfma]; 17[2] August 1601 12 January 1665) was a French lawyer at the Parlement of Toulouse, France, and a mathematician who is given credit for early developments that led to infinitesimal calculus, including his technique of adequality. In particular, he is recognized for his discovery of an original method of finding the greatest and the smallest ordinates of curved lines, which is analogous to that of the differential calculus, then unknown, and his research into number theory. He made notable contributions to analytic geometry, probability, and optics. He is best known for Fermat's Last Theorem, which he described in a note at the margin of a copy of Diophantus' Arithmetica.Pierre de Fermat developed the technique of adequality (adaequalitas in Latin) to calculate maxima and minima of functions, tangents to curves, area, center of mass, least action, and other problems in mathematical analysis. According to Andr Weil, Fermat "introduces the technical term adaequalitas, adaequare, etc., which he says he has borrowed from Diophantus. As Diophantus V.11 shows, it means an approximate equality, and this is indeed how Fermat explains the word in one of his later writings." (Weil 1973).[1] Diophantus coined the word (parisots) to refer to an approximate equality.[2] Claude Gaspard Bachet de Mziriac translated Diophantus's Greek word into Latin as adaequalitas.[citation needed] Paul Tannery's French translation of Fermats Latin treatises on maxima and minima used the words adquation and adgaler.[citation needed]Fermat used adequality first to find maxima of functions, and then adapted it to find tangent lines to curves.To find the maximum of a term , Fermat equated (or more precisely adequated) and and after doing algebra he could cancel out a factor of and then discard any remaining terms involving To illustrate the method by Fermat's own example, consider the problem of finding the maximum of . Fermat adequated with . That is (using the notation to denote adequality, introduced by Paul Tannery):

Canceling terms and dividing by Fermat arrived at

Removing the terms that contained Fermat arrived at the desired result that the maximum occurred when .Fermat also used his principle to give a mathematical derivation of Snell's laws of refraction directly from the principle that light takes the quickest path

B)Method 1 If the quadratic is in the form y = ax2 + bx + c

Example f(x)F(x)=x^2+x+1A=1 b=1 c=1a>0 , therefore y is a minimum value

Yminymin=c-b^2/2aymin=1-1^2/2ymin=3/4

ExampleMethod 2F(x)=x^2+x+1F(x)=(x-1/2)^2-1/4+zF(x)=(x-1/2)^2+3/4Minimum value =3/4

Method3F(x)=x^2+x+1Dy/dx=2ax+bDy/Dx=2X+1Dy/Dx=0 (because minimum value gradient = 0)2X+1=0X=1/2Substitute x into equationY=3/4

ConclusionI never know there are so many ways to find the maximum value of a quadratic equation especially the differentiate part . It is happy to find the more uses of differentiating . In these 3 methods , they are equally useful , I may want to try to use the differentiating method because it could help me master differentiate topic .

xPart 2(a)

y

Length ,xWidth ,yPerimeter(2X+4y)Area , m^2

547.5200237.5

1045200450

1542.5200637.5

2040200800

2537.5200937.5

30352001050

3532.52001137.5

40302001200

4527.52001237.5

50252001250

5522.52001237.5

60202001200

6517.52001137.5

70152001050

7512.5200937.5

8010200800

857.5200637.5

905200450

952.5200237.5

10002000

From this table , we can see dimension that can maximize the barn is 50 m x 25 m which gives total area of 1250 mOr 2X + 4 y = 200 A= XY dy/dx= -X + 50 Y=25 Y=(200 2 X ) / 4 (1) =((200 2 X ) / 4 ) X 0 = -X + 50 = ( -2X^2 + 200X ) / 4 X=50 = - X^2 +50X sub x into equation (1)

So in this question two methods can be use .

hb)

30-2h30-2h

Volume=length x width x height = (30-2h) x (30-2h) x h = h(4h^2 120h + 1900) = 4h ^3 120 h^ 2 + 900 hDifferentiate the volumeDv/Dh = 12 h ^2 240 h + 900Let Dv/ Dh = 012 h ^2 240 h + 900=0H^2-20 h + 75 = 0(h-15) ( h 5) = 0H = 15 ( rejected ) h =5 ( accepted )

VolumeSubstitute h = 5 into V = = 4h ^3 120 h^ 2 + 900 hV=4(5)^3-120(5)^2 + 900 (5)V max = 2000 cm ^ 3

Part 3 P(t)=-1800 cos ( n/6 x t) + 1800P(t) = visitor , t= time1.graph function of p(t)t0930103011301230133014301530163017301830193020302130223023300030

0123456789101112131415

P(t)0241.2900180027003359360033582699179989924002429011802

II)The peak hour reach at 1530 / 3.30 p.m , number of visitor on that hour are 3600 visitor

III) AT 7. 30 P.M , T=10P(10) = -1800 cos ( n/6 x 10 ) + 1800P(10) = 900 Visitor

IV)P(t) = 2570-1800 cos ( n/6 x T ) + 1800 = 2570-1800 cos ( n /6 x T) = 770Cos ( n/6 x T) = -77/180n/6 x T = 2T = 2 x 6 / nT = 3.84 hours3.84 hours = 3 hour + ( 0.84 x 60)=3 hour 50 min0930 + 3 hour 50 min = 1320

Linear programmingLinear programming (LP; also called linear optimization) is a method to achieve the best outcome (such as maximum profit or lowest cost) in a mathematical model whose requirements are represented by linear relationships. Linear programming is a special case of mathematical programming (mathematical optimization).More formally, linear programming is a technique for the optimization of a linear objective function, subject to linear equality and linear inequality constraints. Its feasible region is a convex polytope, which is a set defined as the intersection of finitely many half spaces, each of which is defined by a linear inequality. Its objective function is a real-valued affine function defined on this polyhedron. A linear programming algorithm finds a point in the polyhedron where this function has the smallest (or largest) value if such a point exists.Linear programs are problems that can be expressed in canonical form:

where x represents the vector of variables (to be determined), c and b are vectors of (known) coefficients, A is a (known) matrix of coefficients, and is the matrix transpose. The expression to be maximized or minimized is called the objective function (cTx in this case). The inequalities Axb and x 0 are the constraints which specify a convex polytope over which the objective function is to be optimized. In this context, two vectors are comparable when they have the same dimensions. If every entry in the first is less-than or equal-to the corresponding entry in the second then we can say the first vector is less-than or equal-to the second vector.Linear programming can be applied to various fields of study. It is used in business and economics, but can also be utilized for some engineering problems. Industries that use linear programming models include transportation, energy, telecommunications, and manufacturing. It has proved useful in modeling diverse types of problems in planning, routing, scheduling, assignment, and design.

HistoryThe problem of solving a system of linear inequalities dates back at least as far as Fourier, who in 1827 published a method for solving them,[1] and after whom the method of FourierMotzkin elimination is named.The first linear programming formulation of a problem that is equivalent to the general linear programming problem was given by Leonid Kantorovich in 1939, who also proposed a method for solving it.[2] He developed it during World War II as a way to plan expenditures and returns so as to reduce costs to the army and increase losses incurred by the enemy. About the same time as Kantorovich, the Dutch-American economist T. C. Koopmans formulated classical economic problems as linear programs. Kantorovich and Koopmans later shared the 1975 Nobel prize in economics.[1] In 1941, Frank Lauren Hitchcock also formulated transportation problems as linear programs and gave a solution very similar to the later Simplex method;[2] Hitchcock had died in 1957 and the Nobel prize is not awarded posthumously.During 1946-1947, George B. Dantzig independently developed general linear programming formulation to use for planning problems in US Air Force. In 1947, Dantzig also invented the simplex method that for the first time efficiently tackled the linear programming problem in most cases. When Dantzig arranged meeting with John von Neumann to discuss his Simplex method, Neumann immediately conjectured the theory of duality by realizing that the problem he had been working in game theory was equivalent. Dantzig provided formal proof in an unpublished report "A Theorem on Linear Inequalities" on January 5, 1948.[3] Postwar, many industries found its use in their daily planning.Dantzig's original example was to find the best assignment of 70 people to 70 jobs. The computing power required to test all the permutations to select the best assignment is vast; the number of possible configurations exceeds the number of particles in the observable universe. However, it takes only a moment to find the optimum solution by posing the problem as a linear program and applying the simplex algorithm. The theory behind linear programming drastically reduces the number of possible solutions that must be checked.The linear-programming problem was first shown to be solvable in polynomial time by Leonid Khachiyan in 1979, but a larger theoretical and practical breakthrough in the field came in 1984 when Narendra Karmarkar introduced a new interior-point method for solving linear-programming problems.

A Diet ProblemSuppose the only foods available in your local store are potatoes and steak. The decision about how much of each food to buy is to made entirely on dietary and economic considerations. We have the nutritional and cost information in the following table:Per unit of potatoes Per unitof steak Minimum requirements

Units of carbohydrates 318

Units of vitamins 4319

Units of proteins 137

Unit cost2550

The problem is to find a diet (a choice of the numbers of units of the two foods) that meets all minimum nutritional requirements at minimal cost. a. Formulate the problem in terms of linear inequalities and an objective function.b. Solve the problem geometrically.c. Explain how the 2:1 cost ratio (steak to potatoes) dictates that the solution must be where you said it is.d. Find a cost ratio that would move the optimal solution to a different choice of numbers of food units, but that would still require buying both steak and potatoes.e. Find a cost ratio that would dictate buying only one of the two foods in order to minimize cost.a) We begin by setting the constraints for the problem. The first constraint represents the minimum requirement for carbohydrates, which is 8 units per some unknown amount of time. 3 units can be consumed per unit of potatoes and 1 unit can be consumed per unit of steak. The second constraint represents the minimum requirement for vitamins, which is 19 units. 4 units can be consumed per unit of potatoes and 3 units can be consumed per unit of steak. The third constraint represents the minimum requirement for proteins, which is 7 units. 1 unit can be consumed per unit of potatoes and 3 units can be consumed per unit of steak. The fourth and fifth constraints represent the fact that all feasible solutions must be nonnegative because we can't buy negative quantities. constraints: {3X + Y 8, 4X+ 3Y 19, X+ 3Y 7, X 0, Y 0}; Next we plot the solution set of the inequalities to produce a feasible region of possibilities. c) The 2:1 cost ratio of steak to potatoes dictates that the solution must be here since, as a whole, we can see that one unit of steak is slightly less nutritious than one unit of potatoes. Plus, in the one category where steak beats potatoes in healthiness (proteins), only 7 total units are necessary. Thus it is easier to fulfill these units without buying a significant amout of steak. Since steak is more expensive, buying more potatoes to fulfill these nutritional requirements is more logical. d) Now we choose a new cost ratio that will move the optimal solution to a different choice of numbers of food units. Both steak and potatoes will still be purchased, but a different solution will be found. Let's try a 5:2 cost ratio. d) Now we choose a new cost ratio that will move the optimal solution to a different choice of numbers of food units. Both steak and potatoes will still be purchased, but a different solution will be found. Let's try a 5:2 cost ratio. d) Now we choose a new cost ratio that will move the optimal solution to a different choice of numbers of food units. Both steak and potatoes will still be purchased, but a different solution will be found. Let's try a 5:2 cost ratio. Thus, the optimal solution for this cost ratio is buying 8 steaks and no potatoes per unit time to meet the minimum nutritional requirements.

A Blending ProblemBryant's Pizza, Inc. is a producer of frozen pizza products. The company makes a net income of $1.00 for each regular pizza and $1.50 for each deluxe pizza produced. The firm currently has 150 pounds of dough mix and 50 pounds of topping mix. Each regular pizza uses 1 pound of dough mix and 4 ounces (16 ounces= 1 pound) of topping mix. Each deluxe pizza uses 1 pound of dough mix and 8 ounces of topping mix. Based on the past demand per week, Bryant can sell at least 50 regular pizzas and at least 25 deluxe pizzas. The problem is to determine the number of regular and deluxe pizzas the company should make to maximize net income. Formulate this problem as an LP problem. Let X and Y be the number of regular and deluxe pizza, then the LP formulation is: Maximize X + 1.5 YSubject to:X + Y 1500.25 X + 0.5 Y 50X 50Y 25X 0, Y 0

b) Further exploration Cabinet XCabinet Y

Cost (RM)100200

Space(M^2)0.60.8

Volume ( M^2) 0.81.2

X:2 Y:3X/Y 2/3Y 3 /2 X

100x + 200y < 1400X + 2Y 14 0.6X + 0.8Y 7.23X + 4Y 36

Graph

II) Volume = 0.8 X + 1.2 yWe need to find the value of x and y that maximize the space of cabinet First method = simultaneous equationLet x + 2y = 14 (2) 3x + 14 y = 36 (1) X = 14 2y (3)Substitute (3) into (1) Substitute (4) into ( 3)3(14 2y) + 4y = 36 x= 14 2 (3)42 6y + 4y = 36 x= 82y = 6Y =3 (4) Then substitute x = 8 , y = 3 into V = 0.8 x + 1.2 yArea = 0.8(8) + 1.2 (3) = 10 m^3

Second method ( linear Programming)

Then substitute x = 8 , y = 3 into V = 0.8 x + 1.2 yArea = 0.8(8) + 1.2 (3) = 10 m^3III) Area = 0.6 X + 0.8 Y Cost = 100 x + 200y Volume = 0.8 x + 1.2 y

CombinationXYArea(m^2)Volume ( m^2)Cost (RM)

A456.49.21400

B546.28.81300

C 646.89.61400

D736.69.21300

E837.210.01400

F927.09.61300

IV) If I were Aaron , I would chose combination E ( x = 8 , y = 3 ) because the cost RM 1400 and it also maximize the area 7.2m ^ 2 and volume 10 m ^ 3 with high number of combination of cabinet 11

Conclusion I learn a lot from doing this project especially in linear programming . After learning linear programming , I learnt there are actually ways to help in deciding my daily option . Linear programming could help me decide the best choice under the best condition . I also learnt that nowadays there are technology called graphic calculator that can help generate a graph by just typing some few equation . If I did not do this project , I will never know the existence of this fascinating creation . It could help me solve some graphic question in both math and add maths .I also learned the history and hard works of mathematician in investigating to find all these ways to solve question. Hence , I find the usefulness of internet. A lot of information can be found through internet. In this project , I feel like I am an investigator to find treasure on the internet .

Resourcehttps://en.wikipedia.org/wiki/Pierre_de_Fermathttps://en.wikipedia.org/wiki/Adequalityhttp://www.meta-calculator.com/online/http://www.wikihow.com/Find-the-Maximum-or-Minimum-Value-of-a-Quadratic-Function-Easilyhttp://home.ubalt.edu/ntsbarsh/opre640a/partviii.htm#rdietproblemhttps://en.wikipedia.org/wiki/Linear_programming

Page | 3