Form 4 Mathematics – LINEAR PROGRAMMING msomimaktaba, November 13, 2018February 13, 2019 SIMULTANEOUS EQUATIONS Solving simultaneous equations graphically, the solution is given by the point of intersection of the two lines. Examples 1. Solve graphically the following simultaneous equations. 2x – y = 1 3x+3y =6 Equation 1 2x – y = 1 X intercept, y= 0 2x – y = 1 2x – 0 = 1 x = (, 0) Y intercept, x = o 2x – y = 1 y = 1 (0, -1) Equation 2. 3x + 3y = 6 X intercept,Y=0 3x-0 = 6 X = 2 (2,0) Y intercept,X=0 0 + 3y = 6 3Y = 6 y = 2 (0,2) Solution ( 1,1) i.e x = 1 and y = 1. 2. Ali paid 34 shillings for 10 oranges and 35 mangoes. Moshi went to the same fruit market and paid 24 shillings for 16 oranges and 18 mangoes . what was the price of a mango and for an orange. Solution: Let X be a price of an orange Let y be the price of mangoes 10X + 35 Y = 34 16X + 18 Y = 24 Solution: 380 y = 314 Y = 0.8 10X + 35y = 34 10X + 35 x 0.8 = 34 10X + 28 = 34 X= 0.6 X intercept,Y=0 10 X + 35 Y = 34 10X + 0 = 34 10 X = 34 X = 3.4 (3.4,0) Y intercept , X=0 0 + 35 Y = 34 35Y = 34 Y = 1 (0,1) X intercept Y =0 16X + 18 Y = 24 16X = 24 X = 1.5 (1.5, 0) Y intercept X= 0 18Y = 24 Y = 1.3* (0 ,1.3) x = 0.6 y = 0.8 ...Price of an orange is 0.6sh and price of a mango is 0.8sh. 3. 7 X + 3Y + 12 = 0 5X – 2Y + 2 = 0 Solution: 7X + 3Y + 12 = 0 X intercept y=0 7X + 0 = -12 X = Y intercept X=0 0 + 3Y = -12 3Y = -12 Y = 4 5X – 2Y = -2 X intercept y=0 5X- 0 = -2 X= Y intercept X= 0 0 – 2y = -2 Y = 1 4. 2c + 2d = c – d 2d + 2 = c + 1 Solution: 2c + 2d = c – d 2c –c + 2d + d = 0 c + 3d =0…….(1) 2d + 2 = c + 1 2d + 2- c-1= 0 2d – c + 1= 0 ……..(2) c + 3d = 0 c intercept, d = 0 c = 0 d intercept , c = 0 3d = 0 d = 0 2d – c = -1 C intercept, d= 0 -c = -1 c = 1 d intercept , c= 0 0 – 2d = 1 d = – d = (0, – 0.5) Solution (0.6, -0.2) By substitution method c + 3d = 0 ——-(i) c – 2d = 1——–(ii) equation; c = -3d c -2d = 1 c – 2(-0.2) = 1 c + 0.4 = 1 – 0.4 c = 0.6 :. c = 0.6 , d = -0.2 LINEAR INEQUALITIES When a linear equation of the form ax + by + c = 0 is represented on a coordinate plane, it separates the plane into two disjoint sets.E.g. y = 4 separate the plane into two disjoint setsThe points above the line y= 4 satisfy the relation y>4, while those in the lower half plane satisfy the relation y < 4. SHADING OF REGIONSWhen drawing the inequality y > 4, first draw the line of separation to separate the plane y>4, and the plane y< 4 which is y = 0, Then shade the unwanted region. Note: The shaded region together with the line y = 4 represents y > 4 Boundaries of half- planeEquations are used to describe the boundaries of half- plane. The boundary lines are continuous to include points on the line when the inequality is written using the sign > or < where as the boundary lines are dotted if the inequality is written in the of form < or >.E.g. Draw and show the half- plane representing the inequality 4x + 2y > 8Feasible RegionFeasible region is the region of intersection of all the inequality given in a problem. In order the feasible region to be seen clearly the unwanted region of the inequalities should be shaded leaving the required region clear.Example 1.Show the feasible region which satisfy the following inequalities 3x + 3y > 12 and y – x < 2Solution 3x + 3y > 12, (0, 4) (4, 0) y – x < 2, (0, 2) (-2, 0)From 3x + 3y > 12, after drawing the line of separation i.e 3x + 3y = 12, The point above or below the line is used to test which region satisfy the inequality.e.g. (0, 0) the origin 3(0) + 3(0) > 12 0> 12 Not true... The region below the line does not satisfy the inequality instead. The opposite region satisfies.The region below is shaded to live the wanted region clear. For the case of y – x < 2testing using the origin (0, 0) 0 – 0 < 2 0 < 2 (it is true)The region below is the wanted region, therefore we shade the region above. Example 2A bread dealer can buy up to 150 loaves if bread . premium bread costs 200/= per loaf and royal bread costs 280/= per loaf. The dealer can spend no more than 36000/= in the business. Premium bread sells at a profit of 40/= per loaf while royal bread sells at a profit of 50% per loaf . how many loaves of bread of each type should the dealer buy in order to generate maximum profit? Solution: Let X be the no. of premium bread. Let Y be the no of Royal bread 200X + 250Y ≤ 36000 X + Y ≤ 150 X ≥ 0 Y ≥ 0 X intercept, Y=0 200X+ 250Y = 36,000 200X = 36000 X = 180 (180,0) X intercept, Y = 0 X + Y = 150 X = 150 (150, 0) Y intercept, X = 0 0 + 250 Y = 36,000 250Y = 36,000 Y = 144 ( 0, 144) Y intercept, X = 0 0 + Y = 150 Y = 150 objective function 40X + 50/100 x 250Y = Maximum profit 40X + 125 Y = Maximum profit A (6, 0) B (144, 0) C (0,144) 40X + 125Y = Maximum profit 40 + 125X144 =18,040 The Corner points A (0, 0) C (30, 120) B (0, 144) D (150, 0) f(x, y) = 40x + 125y maximize f(0, 0) = 40x + 125y = 0f(0, 144) = 40x + 125y = 18000f(30, 144) = 40x + 125y = 15120f(150, 0) = 40x + 125y = 6000 optimal point is (0, 144)... In order for the bread dealer to get maximum profit, he should prepare 144 royal breads only. Example 3.Draw a graph and show the feasible region which is satisfied by the inequalities ; X ≥ 0, 2x + y ≤ 4, 2x +3y ≥ 8 , y ≥ 0 , y+ 3x = 9 , x + y ≤ 10 Which If any of the inequalities can be omitted without affecting the answer. SOLUTION 2x + y = 4 X intercept, Y = 0 2x + 0 =4 x = 2 (2, 0) Y intercept 0 + y = 4 y = 4 (0, 4) 2x + 3y = 8 X intercept, Y =0 2x + 0 = 8 2x = 8 x =4 (4, 0) y intercept, x =0 0 + 3Y = 8 y =8/3 (0 , 2.7) x + 3y = 9 x intercept, y =0 x + 0 = 9 x = 9 y intercept, y = 0 0 + 3Y = 9 y = 3 (0, 3) x + y = 10 x intercept, y =0 x =10 (10, 0) y intercept, x =0 y = 10 (0, 10 ) Example 4. George buys X pencils 10 shillings each,( X+8) exercise books at 10 shillings each. If he wishes to have some change as town bus fare from a 200shs note, form an inequality in X and solve it to find the range of X. Solution: X no. of pencils 0 X + 10(X+8) ≤200 10X + 10X + 80 = 200 20 X = 120 X = 6 X < 6 The range of x is = 1, 2,3,4,5. Exercise 8.11. Show by shading out the unwanted regions, the half planes representing the following simultaneous inequalities.a) y < 2x – 1, y > 3 – xb) -3 < x – y<2c) y< 2x, y > 3 – x, y > -1 Solution a) y < 2x – 1 y > 3 – x 1 < 2x – y y + x > 3 2x – y> 1 y + x = 3 2x – y = 1 (0, 3) (3,0)(0, -1) (1/2, 0) (b) -3 < x – y < 2 -3 < x – y x – y < 2 x – y > -3 x – y = 2 x – y = -3 (0, -2) ( 2, 0)(0, 3) (-3, 0) (c) y > 2x – 1 y > 3 – x y – 2x > -1 y + x > 3 y – 2x = -1 y + x = 3 (0, -1) (1/2, 0) (0, 3) (3, 0) y > -1 y = -1 Linear ProgrammingLinear programming is a branch of mathematics which enables to solve problems which one wants to get the greatest or least value of a quantity. Solving linear programming problems1.A farm is to be planted with wheat and maize while observing the following. If wheat yields a profit of 800 Sh. per hectare while maiza yields 600sh.per hectare.How should the area be planted in order to get maximum profit? SolutionStep1:Express the information provided in mathematical formLet x represents number of hectares of wheat planted y represents number of hectares of maize to be planted Step2: The objective functionThis is the function to be optimized. The objective of the problem is to maximize the profit. Therefore 800x – profit yielded by wheat 600y – profit yielded by maizeThe objective function f(x,y) = 800x + 600y is to be maximized. Step3.Availability.2x + y < 10 (Available days of labor)700x + 600y < 4200 (Available money for labors cost)300x + 400y < 2400 (Available money for fertilizer)x > 0, y>0 hectares of wheat and maize can only be positive. (non – negative)... The problem is to maximize f(x,y) = 800x + 600y subject to 2x + y < 10 700x + 600y < 4200 i.e 7x + 6y < 42 300x + 400y < 2400 i.e 3x + 4y < 24 x > 0, y>0Step4.Plot the constraints on the graph 2x + y < 10 (0,10 ) (5,0 ) 7x + 6y < 42 (0,7 ) (6,0 ) 3x + 4y < 24 (0,6 ) (8, 0) step 5.The Identify the corner points.A(0, 6), B(0,0), C(5,0), D(3.3, 2.4), E(2.4, 4.2) Step6.Substitute the corner points to the objects function.f(x, y) = 800x + 600y f(0, 6) = 800x + 600y = 3600f(0, 0) = 800x + 600y = 0f(5, 0) = 800x + 600y = 40000f(3.8, 2.4) = 800x + 600y = 4480f(2.4, 4.2) = 800x + 600y = 4500 Note: The point gives the best solution is called optimal point and the best solution is called the optimal solution... f(2.4, 4.2) is the optimal point and 4500 is the optimal solution.2.4 hectares should be planted wheat and while 4.2 hectares should be planted maize in order to get maximum profit EXERCISE 8.2 1. A doctor prescribes that in order obtain an adequate supply of Vitamins A and C, his patient shall have portions of food 1 and 2. The number of the units of vitamin A and C are given in a table below, The doctor prescribes a minimum of 14 units of vitamin A and 21 units of vitamin C. What are the least number of portion of food 1 and 2 that will fit the doctors prescription? SolutionLet x represents number of portion of food 1 and y represent number of portion of food 2. 3x + y > 14 2x + 7y > 21 x > 0 y > 0Objective functionTo minimize f(x,y) = x + yTo represents the constraint on the x-y plane 3x + y = 14 (0, 14) (4.667, 0) 2x +7y = 21 (0, 3) (10.5, 0) x = 0 y = 0 Corner pointsA(0, 14) B(4.03, 1.9) C(10.5, 0) f(x,y) = x + y f(0, 14) = 0 + 14 f(4.03, 1.9) = 4.03 + 1.9 f(10.5, 0) = 10.5 + 0 Best solution(optimal solution) = 5.93 and the optimal point is (4.03, 1.9).... 4.03 is the portion of the food 1 and 1.9 is the portion of food 2 that should be taken to fit the doctors prescription. Basic Mathematics Study Notes Form 4 Basic Mathematics Study Notes Msomi Maktaba All Notes FORM 4MATHEMATICS
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