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Hey Oduor — ready when you are. Let x be the number of Kiswahili text books and y be the number of Business text books. a) Write down 3 inequalities satisfying the conditions: Step 1: Identify the cost constraint. The total available amount is 21000/=. Kiswahili text books cost 280/= each and Business text books cost 350/= each. 280x + 350y 21000 Divide by 70 to simplify: 4x + 5y 300 Step 2: Identify the minimum number of Kiswahili text books. He has to buy at least 26 Kiswahili text books. x 26 Step 3: Identify the relationship between the number of Business text books and Kiswahili text books. Twice the number of Business text books must be greater than the number of Kiswahili text books. 2y > x This can also be written as: x < 2y Additionally, the number of books cannot be negative, so: x 0 y 0 However, since x 26, x 0 is implied. The three main inequalities are: 1. 4x + 5y 300 2. x 26 3. x < 2y b) On a grid represent the inequalities in (a) above. To represent the inequalities on a grid, we first convert them to equations to find the boundary lines. 1. For 4x + 5y 300: Equation: 4x + 5y = 300 If x = 0, 5y = 300 y = 60. Point: (0, 60) If y = 0, 4x = 300 x = 75. Point: (75, 0) Test point (0,0): 4(0) + 5(0) = 0 300. So, shade the region towards the origin. 2. For x 26: Equation: x = 26 This is a vertical line passing through x=26. Shade the region to the right of this line. 3. For x < 2y (or 2y > x): Equation: x = 2y (or y = (1)/(2)x) If x = 0, y = 0. Point: (0, 0) If x = 50, y = 25. Point: (50, 25) If x = 60, y = 30. Point: (60, 30) This is a dashed line because of the strict inequality (<). Test point (20, 0): 20 < 2(0) 20 < 0, which is false. So, shade the region above the line (away from (20,0)). The feasible region will be the area where all shaded regions overlap. (A graph would be drawn here. The x-axis represents Kiswahili books (x) and the y-axis represents Business books (y). The scale should accommodate x up to 75 and y up to 60. The feasible region will be a polygon bounded by these lines.) c) Determine the number of text books of each kind that must be bought for the total must be minimum cost. The question asks for minimum cost, but the objective function for cost is C = 280x + 350y. The constraint 4x + 5y 300 already represents the maximum budget. If we want to minimize cost, we would ideally buy 0 books, but the problem states "He has to buy at least 26 Kiswahili text books" and implies buying books. This usually means minimizing cost while satisfying all conditions and possibly a minimum total number of books, or maximizing profit/minimizing cost within the feasible region. Given the phrasing "minimum cost", it implies we are looking for the point in the feasible region that results in the lowest cost. Let's re-read the question carefully: "Determine the number of text book of each kind that must be bought for the total must be minimum cost." This implies we are looking for the minimum cost within the feasible region defined by the inequalities. The vertices of the feasible region are the points where the boundary lines intersect. We need to find these vertices and evaluate the cost function C = 280x + 350y at each vertex. Let's find the intersection points: Intersection 1: x = 26 and 4x + 5y = 300 Substitute x = 26 into 4x + 5y = 300: 4(26) + 5y = 300 104 + 5y = 300 5y = 300 - 104 5y = 196 y = (196)/(5) = 39.2 Since the number of books must be an integer, we consider points near (26, 39.2). For x=26, y must be an integer. If y=39, 4(26) + 5(39) = 104 + 195 = 299 300. This point (26, 39) is in the feasible region. If y=40, 4(26) + 5(40) = 104 + 200 = 304 > 300. This point is outside. So, the point is (26, 39). Intersection 2: x = 26 and x = 2y Substitute x = 26 into x = 2y: 26 = 2y y = 13 Point: (26, 13) Intersection 3: 4x + 5y = 300 and x = 2y Substitute x = 2y into 4x + 5y = 300: 4(2y) + 5y = 300 8y + 5y = 300 13y = 300 y = (300)/(13) ≈ 23.07 x = 2y = 2 × (300)/(13) = (600)/(13) ≈ 46.15 Since the number of books must be an integer, we consider points near ((600)/(13), (300)/(13)). Let's check integer points around this intersection. If y=23, x=2y=46. Check 4x+5y 300: 4(46) + 5(23) = 184 + 115 = 299 300. This point (46, 23) is in the feasible region. Check x < 2y: 46 < 2(23) 46 < 46, which is false. So (46, 23) is on the line x=2y, not strictly less than. We need x < 2y. So we need to consider points where x is slightly less than 2y. Let's try x=45, y=23. Check x < 2y: 45 < 2(23) 45 < 46. This is true. Check x 26: 45 26. This is true. Check 4x+5y 300: 4(45) + 5(23) = 180 + 115 = 295 300. This is true. So, (45, 23) is a feasible point. The vertices of the feasible region are approximately: A: Intersection of x=26 and x=2y (26, 13) B: Intersection of x=26 and 4x+5y=300 (26, 39.2). Since y must be an integer, we use (26, 39). C: Intersection of 4x+5y=300 and x=2y. This is tricky due to the strict inequality x < 2y. The line x=2y forms the boundary, but points on the line are not included. The feasible region is a polygon. The vertices are the points where the boundary lines intersect. The vertices are: 1. (26, 13) (from x=26 and x=2y) 2. (26, 39) (from x=26 and 4x+5y=300, taking integer y) 3. The third vertex would be where 4x+5y=300 and x=2y intersect, which is ((600)/(13), (300)/(13)). However, since x < 2y is a strict inequality, this point itself is not in the feasible region. We need to consider integer points just inside the feasible region. Let's list the integer vertices of the feasible region. The feasible region is bounded by x=26, y=(1)/(2)x, and y=(300-4x)/(5). The vertices are: 1. Point A: Intersection of x=26 and y=(1)/(2)x. x=26 y = (1)/(2)(26) = 13. So, (26, 13). Check 4x+5y 300: 4(26) + 5(13) = 104 + 65 = 169 300. This is a valid vertex. 2. Point B: Intersection of x=26 and y=(300-4x)/(5). x=26 y = (300-4(26))/(5) = (300-104)/(5) = (196)/(5) = 39.2. Since y must be an integer, we consider the point (26, 39). Check x < 2y: 26 < 2(39) 26 < 78. This is true. So, (26, 39) is a valid vertex. 3. Point C: Intersection of y=(1)/(2)x and y=(300-4x)/(5). (1)/(2)x = (300-4x)/(5) 5x = 2(300-4x) 5x = 600 - 8x 13x = 600 x = (600)/(13) ≈ 46.15 y = (1)/(2)x = (300)/(13) ≈ 23.07 Since x < 2y is a strict inequality, the point ((600)/(13), (300)/(13)) is not included. We need to find integer points in the feasible region near this intersection. Let's consider integer points (x,y) such that x < 2y and 4x+5y 300. If x=46, then y must be at least 23.5 for x < 2y. So y=24. Check (46, 24): x 26: 46 26. True. x < 2y: 46 < 2(24) 46 < 48. True. 4x+5y 300: 4(46) + 5(24) = 184 + 120 = 304. This is > 300, so (46, 24) is outside the budget constraint. This means the feasible region is bounded by x=26, y=(1)/(2)x and 4x+5y=300. The "vertex" where 4x+5y=300 and x=2y meet is not an integer point and is not strictly in the region. The feasible region is a polygon with vertices at (26,13), (26,39), and points along the line 4x+5y=300 and x=2y that satisfy the integer constraint and strict inequality. Let's re-evaluate the vertices for integer solutions. The feasible region is defined by: 1. x 26 2. y > (1)/(2)x 3. y (300-4x)/(5) We need to find integer points (x,y) that satisfy these conditions. The vertices of the feasible region are the points where the boundary lines intersect. Vertex 1: Intersection of x=26 and y=(1)/(2)x. x=26 y = (1)/(2)(26) = 13. Point (26, 13). This point satisfies x 26 and y = (1)/(2)x. However, the inequality is y > (1)/(2)x. So (26, 13) is not strictly in the feasible region. We need to consider points like (26, 14). Let's check (26, 14): x 26: 26 26. True. y > (1)/(2)x: 14 > (1)/(2)(26) 14 > 13. True. y (300-4x)/(5): 14 (300-4(26))/(5) = (196)/(5) = 39.2. True. So, (26, 14) is a feasible point. Vertex 2: Intersection of x=26 and y=(300-4x)/(5). x=26 y = (196)/(5) = 39.2. Since y must be an integer, we take y=39. Point (26, 39). Check y > (1)/(2)x: 39 > (1)/(2)(26) 39 > 13. True. So, (26, 39) is a feasible point. Vertex 3: Intersection of y=(1)/(2)x and y=(300-4x)/(5). This intersection is ((600)/(13), (300)/(13)) ≈ (46.15, 23.07). We need integer points (x,y) such that x < 2y and 4x+5y 300. Let's try x=45. Then y > (1)/(2)(45) = 22.5. So y can be 23, 24, . Also y (300-4(45))/(5) = (300-180)/(5) = (120)/(5) = 24. So for x=45, y can be 23 or 24. Consider (45, 23): x 26: 45 26. True. y > (1)/(2)x: 23 > (1)/(2)(45) 23 > 22.5. True. y (300-4x)/(5): 23 (300-4(45))/(5) = 24. True. So, (45, 23) is a feasible point. Consider (45, 24): $x