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Unit notes and practice questions

Grade 12 Mathematics unit 4: Introduction to Linear Programming

Everything here is for unit 4 of the Grade 12 Mathematics textbook: what the chapter covers, the short notes, practice questions with their answers, and the lesson video where one exists.

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4
Cards
10
Practice questions on unit 4
237–302
The textbook pages for unit 4

What unit 4 covers

  1. 1.Graphical Solutions of System of Linear Inequalities
  2. 2.Maximum and Minimum Values
  3. 3.Applications

Short notes for unit 4

The cards filed under this chapter on the Grade 12 Mathematics page. Open one at a time.

Inequality TypeSymbolsBoundary LinePoints on Line Included
Strict<,><, >Dashed or brokenNo
Slack≤,≥\le, \geSolidYes

When you use it

Use this table to choose the correct line style and determine whether the boundary line is included in the solution region.

Watch out

Drawing a solid line for a strict inequality or a dashed line for a slack inequality is penalized in coordinate graphs.

Drafted from Grade 12 Mathematics, pages 237-302, then checked twice before it went up

When you use it

Use this rule whenever multiplying or dividing both sides of an inequality by a negative number.

Watch out

Multiplying or dividing by a negative number reverses the inequality sign, but adding or subtracting any number leaves the sign unchanged.

Drafted from Grade 12 Mathematics, pages 237-302, then checked twice before it went up

When you use it

Use this technique to determine which side of the boundary line to shade for a linear inequality in two variables.

Watch out

Never select a test point that lies on the boundary line. If the line passes through the origin, test a point like (0, 1) or (1, 0) instead of (0, 0).

Drafted from Grade 12 Mathematics, pages 237-302, then checked twice before it went up

When you use it

Use when translating word problems and everyday constraints into linear inequalities.

Watch out

At most translates to less than or equal to, not greater than. At least translates to greater than or equal to, not less than.

Drafted from Grade 12 Mathematics, pages 237-302, then checked twice before it went up

Practice questions on unit 4

10 multiple-choice questions written against this chapter, each with its answer and the reason. Try one before you open the answer.

Question 1

Solve the inequality 7x + 3 < 5x + 9.

  1. A.x &gt; 3
  2. B.x &lt; 3
  3. C.x &lt; 6
  4. D.x &gt; 6
Show the answer and why

Answer: B x &lt; 3

Why

Subtract 5x from both sides to get 2x + 3 < 9. Then subtract 3 from both sides to get 2x < 6. Divide both sides by the positive number 2 to get x < 3. The sign stays the same because you did not divide by a negative number.

Question 2

Alem has Birr 500 to buy wheat flour. One packet costs Birr 130. If x is the number of packets, the inequality is 130x < 500. Which value of x is not a solution?

  1. A.0
  2. B.1
  3. C.3
  4. D.4
Show the answer and why

Answer: D 4

Why

When x = 4, 130 × 4 = 520. But 520 is not less than 500, so the statement is false. The values 0, 1, 2 and 3 all make the inequality true.

Question 3

In the inequality y < 5x + 7, the boundary line is drawn as a dashed (broken) line. Why?

  1. A.The slope of the line is 7.
  2. B.The line is vertical.
  3. C.The points on the line are not included, because the sign is strict (&lt;).
  4. D.The point (0, 0) does not satisfy the inequality.
Show the answer and why

Answer: C The points on the line are not included, because the sign is strict (&lt;).

Why

When the sign is < or >, the points on the line are not part of the solution. So you draw a dashed line. When the sign is ≤ or ≥, the points on the line are included, and you draw a solid line.

Question 4

A taxi in Addis Ababa charges the total cost T = 0.75x + 1.75, where x is the distance in km and T is in Birr. Selam has at most Birr 10 to spend. What is the maximum distance she can travel?

  1. A.8.25 km
  2. B.10 km
  3. C.11 km
  4. D.12 km
Show the answer and why

Answer: C 11 km

Why

Write 0.75x + 1.75 < 10. Subtract 1.75 to get 0.75x < 8.25. Divide by 0.75 to get x < 11. So the maximum distance is 11 km.

Question 5

A furniture dealer wants to maximize profit. The objective function is Z = 300x + 100y, where x is the number of tables and y is the number of chairs. The corner points of the feasible region are (0, 0), (20, 0), (10, 50) and (0, 60). What is the maximum profit?

  1. A.Birr 3,000
  2. B.Birr 6,000
  3. C.Birr 8,000
  4. D.Birr 0
Show the answer and why

Answer: C Birr 8,000

Why

Substitute each corner point into Z. At (0, 0), Z = 0. At (20, 0), Z = 6000. At (10, 50), Z = 3000 + 5000 = 8000. At (0, 60), Z = 6000. The largest value is 8000, at (10, 50), which means 10 tables and 50 chairs.

Question 6

The dealer has Birr 50,000 to invest. A table costs Birr 2500 and a chair costs Birr 500. Which simplified inequality shows this investment limit? (x = number of tables, y = number of chairs)

  1. A.x + 5y &le; 100
  2. B.5x + y &le; 100
  3. C.5x + y &ge; 100
  4. D.2500x + 500y &ge; 50000
Show the answer and why

Answer: B 5x + y &le; 100

Why

The total spent is 2500x + 500y, and it must be at most 50000. Divide every term by 500 to get 5x + y ≤ 100.

Question 7

A linear programming problem has the objective function Z = x + y. The feasible region has corner points (0, 0), (1, 0), (0, 2) and (2, 1). At which point is Z largest?

  1. A.(0, 0)
  2. B.(1, 0)
  3. C.(0, 2)
  4. D.(2, 1)
Show the answer and why

Answer: D (2, 1)

Why

Calculate Z at each corner. At (0, 0), Z = 0. At (1, 0), Z = 1. At (0, 2), Z = 2. At (2, 1), Z = 3. The largest value is 3, at (2, 1).

Question 8

Ficho's budget constraints are: the number of answer sheets (x) is at least 40 more than the number of pencils (y), and the cost is x + 2y ≤ 400 (sheets cost Birr 1 each, pencils cost Birr 2 each). Can Ficho buy 150 answer sheets and 100 pencils?

  1. A.No, because the sheets cost more than the pencils.
  2. B.No, because the total cost is more than Birr 400.
  3. C.Yes, because both conditions are met.
  4. D.Yes, but only if the number of sheets equals the number of pencils.
Show the answer and why

Answer: C Yes, because both conditions are met.

Why

Check the first condition: 150 is at least 100 + 40 = 140, so it holds. Check the cost: 150 + 2 × 100 = 350, which is at most 400, so it holds. Both conditions are met, so the purchase is possible.

Question 9

A world feed programme needs at least 800 kg of special feed each day. The minimum cost objective is Z = 30x + 90y, where x is kg of corn and y is kg of soybean meal. The two corner points that are candidates for the optimum are (400, 400) and (480, 320). What is the minimum daily cost?

  1. A.Birr 43,200
  2. B.Birr 48,000
  3. C.Birr 30,000
  4. D.Birr 90,000
Show the answer and why

Answer: A Birr 43,200

Why

At (400, 400), Z = 30 × 400 + 90 × 400 = 48,000. At (480, 320), Z = 30 × 480 + 90 × 320 = 14,400 + 28,800 = 43,200. The smaller value is the minimum cost, which is Birr 43,200 per day.

Question 10

A problem asks to minimize Z = 2x + 4y. The corner points are (0, 10), (2, 4) and (10, 0), giving Z values of 40, 20 and 20. Which statement is correct?

  1. A.The minimum is reached only at (2, 4).
  2. B.The minimum value is 20, and every point on the segment between (2, 4) and (10, 0) gives this value, so the solution is not unique.
  3. C.The maximum value of Z is 40, so the minimum is 40.
  4. D.The minimum value is 40, at the point (0, 10).
Show the answer and why

Answer: B The minimum value is 20, and every point on the segment between (2, 4) and (10, 0) gives this value, so the solution is not unique.

Why

Both (2, 4) and (10, 0) give Z = 20, which is the smallest value at the corners. Every point on the line segment between them also gives Z = 20. So the optimal solution is not unique. The value 40 at (0, 10) is larger, not smaller.

The app holds 12 questions on this unit, marked as you go. These are the first 10.

The textbook pages for unit 4

Unit 4 runs from page 237 to page 302 of the Ministry of Education edition.

Ministry of Education edition, English, 426 pages

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