Chapter II Inequalities

Transcription

1 Chapter II Inequalities Linear and quadratic inequalities in one variable 1 Basic properties of inequalities For any three real numbers a, b and c: Property Example 1. If a>b and b>c, then a>c. 9>5 and 5>, then 9>. If a>b, then a+c>b+c. 3>1, then 3+5> If a>b, then 8>6, then ac>bc when c>0; 8()>6() for >0; ac<bc when c<0. 8(-)<6(-) for <0. 4. If a>b (where ab>0), then 4>, then < < a b 4 5. If a 0, then a > , then ( 3) > 0. How to solve Linear inequality in one variable How to solve ax-b(<,,>, )0? Express the inequality as ax (<,,>, ) b. (Use property ) Use property 3 to solve the inequality 3 Graphical Representations Solution Graph Remark x < 5 Put variable x on LHS. Put number on RHS. Arrow is in the same direction as the -5 0 Inequality sign. x 1 Means -5 is not included in the solution. 0 1 Means 1 is included in the solution. 4 Compound inequalities involving and, or Steps: Solve each inequality separately.

2 Represent each solution graphically. Shape the required region: AND : The common region will give the solution for x. OR : Any region marked in step () will make up the solutions for x. State the final answer. 5 Quadratic inequality in one variable. (1) Express the inequality as ax + bx + c (<,,>, ) 0 with a > 0. () Find the roots of ax + bx + c = 0 and sketch the graph of y = ax + bx + c. No real roots Double root Two distinct roots root root root (3) Read the answer from the graph. Class practice: use step 3 to find the value of x in the following graphs. x 3x >0 x 6x +5 < 0 -x + 6x + 7 > 0 y = x 3x y = x 6x + 5 y = x + 6x + 7 x < 0 or x > < x < < x < 7 Linear inequalities in two variables 1 To sketch the region represented by a linear inequality in two variables: Sketch the straight line obtained by replacing the inequality with an equality. Choose a test point not on the line ((0,0) is a good choice if the line does not pass through the origin, and if the line does pass through the origin a point on one of the axes would be a good choice). If the test point satisfies the inequality, then the set of solutions is the entire region on the same side of the line as the test point. Otherwise it is the region

3 on the other side of the line. In either case, shade out the side that does not contain the solutions, leaving the solution region showing. Example. To sketch the linear inequality 3x-4y 1. Firstly, sketch the line 3x-4y = 1. Next, choose the origin (0, 0) as the test point (since it is not on the line). Substituting x=0, y=0 in the inequality gives 3(0)-4(0) 1. Since this is a true statement, (0, 0) is in the solution set, so the solution set consists of all points on the same side as (0, 0). This region is left unshaded, while the (grey) shaded region is blocked out. Feasible Region The feasible region determined by a collection of linear inequalities is the collection of points that satisfy all of the inequalities. To sketch the feasible region determined by a collection of linear inequalities in two variables: Sketch the regions represented by each inequality on the same graph, remembering to shade the parts of the plane that you do not want. What is unshaded when you are done is the feasible region.

4 Example The feasible region for the following collection of inequalities is the unshaded region shown below (including its boundary). 3x-4y 1, x + y 4 x 1 y 0. Linear Programming Problems 1 Example (The Product Mix Problem): A fertilizer factory manufactures two types of fertilizers, Type A is high in phosphorus and Type B is low in phosphorus. Each has three key ingredients: urea, rock phosphate and potash. These ingredients are also produced or mined by the company at other sites. Table 1 gives the information about the fertilizers. Assuming that the company can sell all the fertilizer it can produce, how much of each type should it make, and how much of each raw material will it need per day? Model:

5 If we know how much of each type we make per day, then it is easy to compute the raw material requirements. Step 1. The decision variables Let us denote the daily production of Type A by x tons, and of Type B by y tons. These variables (whose values we must determine) are called the decision variables. Identifying the decision variables, what type of values they can have, is the first key step in modelling a problem. Here, x and y are real numbers, which must be positive. Step. The objective function We would like to maximize our profit. Based on table 1, our profit per day = 15x + 10y. Thus the objective function is to maximize the value of the function, z = 15x + 10y. Step 3. The constraints Looking at the objective function, it is clear that if we keep increasing the value of x, y, or both, then our profit also keeps increasing. However, we cannot produce arbitrarily large amount of fertilizer, because our raw material supplies are limited. These, therefore, are constraints on our solution. Since the supply of each raw material is limited, we need one constraint equation for each, as follows: Urea: x + y 1500 Potash: x + y 100 Rock Phosphate: x 500. Step 4. Complete the formulation The problem is therefore completely specified as follows: maximize z( x, y) = 15 x + 10y subject to x + y 1500 x + y 100 x 500 x 0 y 0. So we have managed to convert words into algebra; but how do we solve this problem? Standard Maximization Problem A standard maximization problem in n unknowns is a linear programming problem in which we are required to maximize (not minimize) the objective function, subject to constraints of the form x 0, y 0, z 0,..., and further constraints of the form Ax + By + Cz +... N, where A, B, C,... and N are numbers with N nonnegative.

6 Note that the inequality here must be a "," and not "=" or "." 3 Linear Programming (LP) Problem A linear programming problem is one in which we are to find the maximum or minimum value of a linear expression ax + by + cz +... (called the objective function), subject to a number of linear constraints of the form Ax + By + Cz +... N or Ax + By + Cz +... N. The largest or smallest value of the objective function is called the optimal value, and a collection of values of x, y, z,... that gives the optimal value constitutes an optimal solution. The variables x, y, z,... are called the decision variables. 4 Graphical Method The graphical method for solving linear programming problems in two unknowns is as follows. A. Graph the feasible region. B. Compute the coordinates of the corner points. C. Substitute the coordinates of the corner points into the objective function to see which gives the optimal value. D. If the feasible region is not bounded, this method can be misleading: optimal solutions always exist when the feasible region is bounded, but may or may not exist when the feasible region is unbounded. The textbook shows a straightforward way for determining whether optimal solutions exist in the case of unbounded feasible regions. Example Minimize C = 3x + 4y subject to the constraints 3x 4y 1, x + y 4 x 1, y 0. The feasible region for this set of constraints was shown above. Here it is again with the corner points shown.

7 The following table shows the value of C at each corner point: Therefore, the solution is x = 1, y = 1.5, giving the minimum value C = 9.