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TH EDITION LIAL HORNSBY SCHNEIDER COLLEGE ALGEBRA

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Systems of Inequalities and Linear Programming Solving Linear Inequalities Solving Systems of Inequalities Linear Programming

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Solving Linear Inequalities A linear inequality in two variables is an inequality of the form where A, B, and C are real numbers, with A and B not both equal to 0. (The symbol could be replaced with ,.) The graph of a linear inequality is a half-plane, perhaps with its boundary.

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Solving Linear Inequalities For example, to graph the linear inequality first graph the boundary. Since the points of the line satisfy the equation, this line is part of the solution set.

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Solving Linear Inequalities To decide which half-plane (the one above the line 3x – 2y = 6 or the one below the line) is part of the solution set, solve the original inequality for y. Subtract 3x. Divide by − 2. Reverse the inequality symbol when dividing by a negative number.

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Solving Linear Inequalities For a particular value of x, the inequality will be satisfied by all values of y that are greater than or equal to Thus, the solution set contains the half-plane above the line.

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Caution A linear inequality must be in slope-intercept form (solved for y) to determine, from the presence of a symbol, whether to shade the lower or upper half-plane. In the previous slide, the upper half-plane is shaded, even though the inequality is 3x – 2y 6 is (with a < symbol) in standard form. Only when we write the inequality as (slope-intercept form) does the > symbol indicate to shade the upper half-plane.

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Example 1 GRAPHING A LINEAR INEQUALITY Graph x + 4y > 4. Solution The boundary of the graph is the straight line x + 4y = 4. Since points on this line do not satisfy x + 4y > 4, it is customary to make the line dashed. To decide which half-plane represents the solution set, solve for y.

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Example 1 GRAPHING A LINEAR INEQUALITY Subtract x. Divide by 4. Since y is greater than the graph of the solution set is the half-plane above the boundary.

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Example 1 GRAPHING A LINEAR INEQUALITY Original inequality. Use (0, 0) as a test point. Alternatively, or as a check, choose a test point not on the boundary line and substitute into the inequality. The point (0, 0) is a good choice if it does not lie on the boundary, since the substitution is easily done. False

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Example 1 GRAPHING A LINEAR INEQUALITY Original inequality. Use (0, 0) as a test point. False Since the point (0, 0) is below the boundary, the points that satisfy the inequality must be above the boundary, which agrees with the result.

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Graphing Inequalities 1.For a function , the graph of y < (x) consists of all the points that are below the graph of y = (x); the graph of y > (x) consists of all the points that are above the graph of y = (x). 2. If the inequality is not or cannot be solved for y, choose a test point not on the boundary. If the test point satisfies the inequality, the graph includes all points on the same side of the boundary as the test point. Otherwise, the graph includes all points on the other side of the boundary.

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Solving Systems of Inequalities The solution set of a system of inequalities, such as is the intersection of the solution sets of its members. We find this intersection by graphing the solution sets of all inequalities on the same coordinate axes and identifying, by shading, the region common to all graphs.

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Example 2 GRAPHING SYSTEMS OF INEQUALITIES Solution a. Graph the solution set of each system.

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Example 2 GRAPHING SYSTEMS OF INEQUALITIES Solution b. Graph the solution set of each system.

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Example 2 GRAPHING SYSTEMS OF INEQUALITIES Writing x 3 as – 3 x 3 shows that this inequality is satisfied by points in the region between and including x = – 3 and x = 3.

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Example 2 GRAPHING SYSTEMS OF INEQUALITIES The set of points that satisfies y 0 includes the points below or on the x-axis.

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Example 2 GRAPHING SYSTEMS OF INEQUALITIES Graph y = x + 1 and use a test point to verify that the solutions of y x + 1 are on or above the boundary.

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Example 2 GRAPHING SYSTEMS OF INEQUALITIES Since the solution sets of y 0 and y x +1 have no points in common, the solution set of the system is ø. The solution set of the system is ø, because there are no points common to all three regions.

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Note While we gave three graphs in the solutions of Example 2, in practice we usually give only a final graph showing the solution set of the system.

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Solving a Linear Programming Problem Step 1 Write the objective function and all necessary constraints. Step 2 Graph the region of feasible solutions. Step 3 Identify all vertices or corner points. Step 4 Find the value of the objective function at each vertex. Step 5 The solution is given by the vertex producing the optimal value of the objective function.

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Example 3 FINDING A MAXIMUM PROFIT MODEL The Charlson Company makes two products— MP3 players and DVD players. Each MP3 gives a profit of $30, while each DVD player produces $70 profit. The company must manufacture at least 10 MP3s per day to satisfy one of its customers, but no more than 50 because of production problems. The number of DVD players produced cannot exceed 60 per day, and the number of MP3s cannot exceed the number of DVD players. How many of each should the company manufacture to obtain maximum profit?

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Example 3 FINDING A MAXIMUM PROFIT MODEL Solution First we translate the statement of the problem into symbols. Let x = number of MP3s to be produced daily, and y = number of DVD players to be produced daily. The company must produce at least 10 MP3s (10 or more), so Since no more than 50 MP3s may be produced,

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Example 3 FINDING A MAXIMUM PROFIT MODEL Solution No more than 60 DVD players may be made in one day, so The number of MP3s may not exceed the number of DVD players translates as The numbers of MP3s and of DVD players cannot be negative, so

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Example 3 FINDING A MAXIMUM PROFIT MODEL Solution These restrictions, or constraints, form the system of inequalities Each MP3 gives a profit of $30, so the daily profit from production of x MP3s is 30x dollars. Also, the profit from production of y DVD players will be 70y dollars per day. Total daily profit is, thus, This equation defines the function to be maximized, called the objective function.

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Example 3 FINDING A MAXIMUM PROFIT MODEL Solution To find the maximum possible profit, subject to these constraints, we sketch the graph of each constraint.

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Example 3 FINDING A MAXIMUM PROFIT MODEL Solution The only feasible values of x and y are those that satisfy all constraints— that is, the values that lie in the intersection of the graphs of the constraints.

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Example 3 FINDING A MAXIMUM PROFIT MODEL Solution Any point lying inside the shaded region or on the boundary satisfies the restrictions as to the number of MP3s and DVD players that may be produced.

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Example 3 FINDING A MAXIMUM PROFIT MODEL Solution (For practical purposes, however, only points with integer coefficients are useful.) This region is called the region of feasible solutions. The vertices (singular vertex) or corner points of the region of feasible solutions have coordinates

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Example 3 FINDING A MAXIMUM PROFIT MODEL We must find the value of the objective function 30x + 70y for each vertex. We want the vertex that produces the maximum possible value of 30x + 70y. Maximum The maximum profit, obtained when 50 MP3s and 60 DVD players are produced each day, will be 30(50) + 70(60) = 5700 dollars per day.

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Example 3 FINDING A MAXIMUM PROFIT MODEL The Charlson Company needed to find values of x and y in the shaded region that produce the maximum profit—that is, the maximum value of 30x + 70y. To locate the point (x, y) that gives the maximum profit, add to the graph lines corresponding to arbitrarily chosen profits of $0, $1000, $3000, and $7000: To show why the point of the feasible solution works:

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Example 3 FINDING A MAXIMUM PROFIT MODEL Each point on the line 30x + 70y = 3000 corresponds to production values that yield a profit of $3000. The region of feasible solutions are shown with these lines. The lines are parallel, and the higher the line, the greater the profit. The line 30x + 70y = 7000 yields the greatest profit but does not contain any points of the region of feasible solutions.

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Example 3 FINDING A MAXIMUM PROFIT MODEL To find the feasible solution of greatest profit, lower the line 30x + 70y = 7000 until it contains a feasible solution- that is, until it just touches the region of feasible solutions. This occurs at point A, a vertex of the region. The result observed here hold for every linear programming problem.

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Fundamental Theorem of Linear Programming If an optimal value for a linear programming problem exists, it occurs at a vertex of the region of feasible solutions.

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Example 4 FINDING A MINIMUM COST MODEL Robin takes vitamin pills each day. She wants at least 16 units of Vitamin A, at least 5 units of Vitamin B 1 and at least 20 units of Vitamin C. She can choose between red pills, costing 10 cents each, that contain 8 units of A, 1 of B 1 and 2 of C; and blue pills, costing 20 cents each, that contain 2 units of A, 1 of B 1 and 7 of C. How many of each pill should she buy to minimize her cost and yet fulfill her daily requirements?

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Example 4 FINDING A MINIMUM COST MODEL Since Robin buys x of the 10 cent pills and y of the 20 cent pills, she gets 8 units of Vitamin A from each red pill and 2 units of Vitamin A from each blue pill. Altogether she gets 8x +y units if A per day. Since she wants at least 16 units,

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Example 4 FINDING A MINIMUM COST MODEL Each red pill and each blue pill supplies 1 unit of Vitamin B 1. Robin wants at least 5 units per day, so For Vitamin C, the inequality is Also, x 0 and y 0, since Robin cannot buy negative numbers of the pills.

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Example 4 FINDING A MINIMUM COST MODEL Step 2 The intersection of the graphs of

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Example 4 FINDING A MINIMUM COST MODEL Step 3 The vertices are (0, 8), (1, 4), (3, 2), and (10, 0).

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Example 4 FINDING A MINIMUM COST MODEL Steps 4 and 5 We find that the minimum cost occurs at (3, 2). PointCost = 10x + 20y (0, 8) 10(0) + 20(8) = 160 (1, 4)10(1) + 204) = 90 (3, 2)1(3) + 20(2) = 70 (10, 0)10(10) + 20(0) = 100 Minimum Robin’s best choice is to buy 3 red pills and 2 blue pills, for a total cost of 70 cents per day. She receives just the minimum amounts of Vitamins B 1 and C, and an excess of Vitamin A.

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