SUBJECT: MATHEMATICS
CLASS: SS 2
DATE:
TERM: 2nd TERM
REFERENCE BOOKS
TOPIC: INEQUALITIES CONTENT -Application of Linear Inequalities in Real Life. -Introduction to Linear Programming. APPLICATION OF LINEAR INEQUALITIES IN REAL LIFE. Greatest and Least Values Example Draw a diagram to show the region which satisfies the following inequalities. 5x + y - 4, x + y 4, y x + 2, y – 2x - 4 Find the greatest and the least value of the linear function F = x + 2y within the region. Solution For the inequality 5x + y - 4, first draw the line 5x + y = - 4. When x = 0, y = 4, when x = -1, y = 1 Add a third point on your own and then draw line 5x + y = -4. You may need to extend the axes to do this: Now use a test point such as x = 0, y = 0 When x = 0, y = 0, then 0 - 4 is true, so shade the region below the line 5x + y = -4. For the inequality x + y 4, first draw the line x + y = 4. When x = 0, y = 4 and when y = 0, x = 4. So draw a line that passes through (0, 4) and (4, 0). Test point: (0, 0), so 0 4 is true. Shade the region above the line. Similarly, for y x + 2 and y – 2x - 4, shade the unwanted regions. The required region is labeled as R as shown. R is also called the feasible region (i.e. the region that satisfies a set of inequalities). The greatest (maximum) and the least (minimum) of any linear function such as F = x + 2y occurs at the vertices (corner points) of the region which satisfies the given set of the inequalities. At A(-1, 1) F = x + 2y F = -1 + 2 = 1 At B(1, 3) F = x + 2y F = 1 + 6 = 7 At C(2.67, 1.33) F = x + 2y F = 2.67 + 2.66 = 5.33 At D(0, -4) F = x + 2y F = 0 – 8 = -8 ∴ F = x + 2y is least at the point D(0, 4). F = x + 2y is greatest at the point B(1, 3). Note: The coordinates at point C can also be found by solving the simultaneous equations x + y = 4 and y – 2x = -4, Which gives x = 83 and y = 43. Linear Programming In many real-life situations in business and commerce there are restrictions or constraints, which can affect decision-making. Typical restrictions might be the amount of money available for a project, storage constraints, or the number of skilled people in a labour force. In this section we will see that problems involving restrictions can often be solved by using the graphs of linear inequalities. This method is called linear programming. Linear programming can be used to solve many realistic problems. Example 1 A student has N500. She buys pencils at N50 each and erasers at N20 each. She gets at least five of each and the money spent on pencils is over N100 more than that spent on erasers. Find a. How many ways the money can be spent, Let the student buy x pencils at N50 and y erasers at N20. From the first two sentences, 50x + 20y 500 5x + 2y < 50 (1) Since she gets at least five of each, x 5 (2) y 5 (3) From the third sentence, 5x – 2y > 10 Inequalities (1), (2), (3) and (4) are shown below Example 2 To start a new transport company, a businessman needs at least 5 buses and 10 minibuses. He is not able to run more than 30 vehicles altogether. A bus takes up 3 units of parking space, a minibus takes up to 1 unit of parking space and there are only 54 units available. If x and y are the numbers of buses and minibuses respectively, x 5 y 10 From the second sentence, x + y 30 from the third sentence, 3x + y 54 EVALUATION GENERAL EVALUATION/REVISION QUESTIONS READING ASSIGNMENT New General Mathematics SSS2, pages 98-111, exercise 10g. WEEKENND ASSIGNMENT Objectives 5.Solve the inequality 4y-7<2(3y-1) A. y < -5/2 B. y> -2/5 C. y< -5/3 D. y> -5/2 Theory Printer Cost Working space Type P N15, 000 4000 cm2 Type Q N25, 000 3000 cm2 The man has 48 000cm2 of working space and he can spend up to N290, 000 to buy these machines.
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