Chapter 2: Geometric Linear Programming Introduction In this chapter and in Chapter 3, ou ll be learning how to solve linear programming problems. These problems show up in a wide variet of areas and, in particular, are often seen in economics and business applications. The solution of a linear programming problem involves finding a maimum (or minimum) value for a linear function such as the profit or cost associated with producing some product. The function is assumed to be subject to certain constraints (such as restrictions on the resources necessar to produce the product) which are epressed as linear inequalities. This chapter concentrates on teaching ou how to solve these sorts of problems using the geometric method (also known as the graphical method or the method of corners) while the method described in Chapter 3 is known as the simple method and is an algebraic method, using matri techniques. Here are the steps ou ll learn to use in this chapter for solving a linear programming problem using the graphical approach. 1. Translate into mathematics a problem which will be stated in everda conversational language. Determine the variables involved, state the function to be maimized or minimized and write the constraint inequalities. While linear programming problems often involve several variables or unknowns, usuall onl two unknowns are used to state a problem that is solved using the graphical technique. 2. Graph the region in the plane that is represented b the set of inequalities. 3. Use the method eplained in Section 3 to solve the problem. 2.1 Graphing Inequalities This section will concentrate on step 2 of the above outline, (graphing regions in the plane represented b inequalities), leaving step 1 for Section 2.2.
42 Chapter 2: Geometric Linear Programming If ou need to review graphing points and lines in the plane and finding points of intersection of lines, ou should read through the following material on graphing straight lines and work the problems at the end of the discussion. Graphing straight lines A straight line is represented algebraicall b a linear equation of the form: a + b = c, where a,b and c are constants. Before graphing the line this equation represents, let s first sketch a set of perpendicular lines, one horizontal and the other one vertical. The horizontal line is usuall called the ais and the vertical line is usuall called the ais. The point of intersection of the two lines is called the origin. Mark off a number scale on each ais, with the positive numbers going to the right of the origin on the ais and the negative numbers going to the left of the origin. On the ais, the positive numbers go above the origin and the negative numbers go below. 2 1-3 -2-1 1 2 3 4-1 -2 A point P in the plane can now be represented uniquel b an ordered pair of real numbers (,). To find the values for and, take the point in the plane and construct perpendiculars to each of the aes. The points of intersection of each of these perpendiculars with the aes have real numbers associated with them; the real number associated with the point on the ais is called the coordinate and the real number associated with the point on the ais is called the coordinate.
Section 2.1: Graphing Inequalities 43 P(,) Of course, given an ordered pair of real numbers (,), the procedure can be reversed to graph the ordered pair in the coordinate sstem. This is done b first locating the point corresponding to on the ais and drawing a line through perpendicular to the ais and then finding the point corresponding to on the ais and drawing a line through perpendicular to the ais. The point of intersection of these two perpendiculars is the graph of the ordered pair (,). (,) The ordered pairs A(3, 2), B( 1,4), C(4, 1) and D( 3, 2) are graphed below.
44 Chapter 2: Geometric Linear Programming B(-1,4) 4 2 A(3,2) -6-4 -2 2 4 6 8 C(4,-1) -2 D(-3,-2) -4 Now that ou ve seen how to graph points, let s go back to the problem of graphing lines. Suppose ou re given the linear equation a + b = c. The graph of this equation is a straight line. One wa to graph a straight line is to find two points that lie on that line (and mabe find a third point as a check for the first two). A point that lies on the line is found b determining an ordered pair of numbers (,) that satisfies the equation and then graphing that ordered pair. The easiest points to graph are usuall the intercepts (the points where the given line intersects the aes). The intercept is the point at which the given line intersects the ais which means the coordinate of the point has to be 0. So if = 0 is substituted in the equation and the equation is solved for, that value of will be the intercept. Similarl, to find the intercept, substitute = 0 into the equation and solve for. Eample 2.1. Graph 3 2 = 6. Solution: Letting = 0 and solving for gives = 2. Letting = 0 and solving for ields = 3. So the intercept is 2 and the intercept is 3. The two points (2,0) and (0, 3) are then plotted on the graph below and the line passing through those two points is drawn.
Section 2.1: Graphing Inequalities 45 4 2-6 -4-2 2 4 6 8-2 -4 Eample 2.2. Graph 2 + 5 = 8. Solution: below. The intercept is 4 and the intercept is 8/5. The line is graphed 4 2-6 -4-2 2 4 6 8-2 -4 The onl problem that occurs with this method of graphing straight lines is if the and intercepts turn out to be the same point, namel the origin. In that case ou have to come up with a second point that lies on the line. That s not too tough just substitute a nonzero value for either or into the equation, solve for the other variable and then graph that ordered pair in addition to the origin. Eample 2.3. 3 5 = 0. Solution: Both the and intercepts turn out to be 0, so the origin is one point on the line. To find another point on the line, let s take = 3. Then = 5.
46 Chapter 2: Geometric Linear Programming 4 2-6 -4-2 2 4 6 8-2 There are, of course, other techniques that can be used to graph lines, such as the point slope method or the slope intercept method, which ou probabl remember from previous courses in mathematics. Those techniques can be used if ou prefer them. Recall that the slope of a line L is defined b m = 2 1 2 1 where ( 1, 1 ) and ( 2, 2 ) are two distinct points on the line and 1 2. If 1 = 2, then the slope of L is said to be undefined. In that case, L is parallel to the ais. Facts about slope 1. Two lines, L 1 and L 2, have the same slope if and onl if the are parallel. 2. Suppose the line L 1 has slope m 1 0 and the line L 2 has slope m 2 0. Then m 1 m 2 = 1 if and onl if L 1 and L 2 are perpendicular. If either of the lines has slope 0 (i.e. the line is parallel to the ais) then the two lines are perpendicular if and onl if the other line has a slope which is undefined. In graphing regions represented b linear inequalities, it s going to be ver important to be able to find the point of intersection of two straight lines. Assume the two straight lines, L 1 and L 2, have equations a 1 + b 1 = c 1 and a 2 + b 2 = c 2. To find the point of intersection of these two lines means it s necessar to find a point that lies on both of the lines and hence satisfies both of the equations. To find an ordered pair of numbers (,) that satisfies both equations, ou can use the matri techniques ou
Section 2.1: Graphing Inequalities 47 learned in Chapter 1 or an of the techniques ou ve learned in previous courses. You ma use the method that is quickest and most accurate for ou. Eample 2.4. Find the point of intersection of the lines: 2 + 3 = 6 2 = 11 Solution: Multipling the second equation b 2 and adding the two equations together gives 2 + 3 = 6 2 + 4 = 22 7 = 28 So = 4 and thus = 3. The point of intersection is (3, 4). Eample 2.5. Find the point of intersection of the lines: Solution: 3 4 = 10 2 + 6 = 10 R 1 + R 2 R 2 + 2R 1 3 4 10 2 6 10 1 2 0 2 6 10 1 2 0 0 10 10 1 10R 2 R 1 2R 2 1 2 0 0 1 1 1 0 2 0 1 1 so the point of intersection is ( 2,1). Problems In Eercises 1-6, graph the straight line. Find the and intercepts and label them on the graph.
48 Chapter 2: Geometric Linear Programming 1. = 2 4 2. 3 5 = 10 3. + 8 = 0 4. 2 = 5 8 5. 3 + 2 = 12 6. 5 = 3 In Eercises 7-12, graph the pairs of straight lines. Determine all points of intersection if an eist. If none do, sa so. 7. 2 3 = 12 8. 3 + 4 = 5 + 4 = 5 2 7 = 13 9. 2 = 4 10. 6 + = 14 2 = 8 3 12 = 2 28 11. 4 + 3 = 18 12. 2 = 3 5 2 = 5 5 = 14 24 In producing a commodit, economists often like to talk about the break-even point. The break-even point is the level of production at which cost = revenue. For eample, if the cost of producing copies of this tet is C = 9 + 15,000 (dollars) and the revenue is R = 12 (dollars), the break-even point is 12 = 9 + 15,000 or = 5000 copies. Geometricall, the break-even point represents the point of intersection of the two straight lines C = 9 + 15,000 and R = 12, graphed on the same set of aes. 60,000 C = 9 + 15,000 (5000,60000) R = 12 20,000 10,000 1000 5000 Note that if 5000 then the revenue line is above the cost line and a profit is made. If < 5000, then the revenue line is below the cost line and a loss is incurred. Use these ideas in doing Eercises 13 and 14.
Section 2.1: Graphing Inequalities 49 13. A lumber mill charges $1.50/board foot for oak boards. The cost of producing the boards is $100/da in fied costs and $.25/board foot in variable costs. (a) Sketch the graph of the equations representing the revenue and the cost (on the same set of aes). (b) How man board feet should the mill produce each da to break even? (c) Sketch the graph of the equation representing the profit. (d) At what point does the graph of the profit equation cross the ais? Interpret our result. 14. A woman decides to bu a knitting machine for $1500. If she can make one sweater in two hours and decides she ought to be paing herself $10 per hour for her knitting, how much should she sell the sweaters for (above the cost of the arn) if she is willing to work 20 hours per week and wishes to break even b the end of 30 weeks of work? Graphing Linear Inequalities Graphing linear inequalities in the plane will involve first graphing straight lines. Graphing linear inequalities In solving linear programming problems using the graphical technique, itʼs necessar to graph accuratel. You need to: 1. Make sure the aes are straight and perpendicular. The also have to be of reasonable size and, in general, itʼs a good idea to make them at least 4 inches long. If ou canʼt draw a straight line freehand, use a ruler. 2. Make sure the number scale is marked off uniforml on each ais. In other words, the distance between our basic units ought to be the same from one unit to another. Again, if ou canʼt do it freehand, use a ruler. 3. If ou reall have trouble graphing, use graph paper and a ruler. Suppose the problem is to graph the region of the plane that is represented b the linear inequalit
50 Chapter 2: Geometric Linear Programming a + b < c That means finding all points (,) whose coordinates satisf the inequalit. Now the graph of the set of all points (,) whose points satisf the equalit a + b = c is a straight line. This line divides the plane into two half planes. One of the half planes represents the set of points given b the inequalit a + b < c and the other half plane represents the set of points given b the inequalit a + b > c a + b = c (points on the line) a + b < c (points in one half-plane) a + b > c (points in other half-plane) Thus to graph a given inequalit, first graph the corresponding equation and then figure out which half plane represents the inequalit. The easiest wa to do that is pick a point in one of the half planes (NOT on the line) and see if the coordinates of that point satisf the inequalit. If the do, ou ve got the correct half plane; if not, ou want the other one. In picking the point, ou might as well pick an eas one to check the origin is alwas one of the easiest provided it doesn t lie on the line between the two halfplanes. In an case, if the inequalit is of the or tpe, the line is included in the region and is represented on the graph as a solid line. If the inequalit is of the < or > tpe, the line is not included and is represented on the graph as a dotted line.
Section 2.1: Graphing Inequalities 51 Eample 2.6. Graph the region represented b the inequalit 2 3 6 Solution: 4 2 2 3 < 6 2 3 = 6-6 -4-2 2 4 6 8-2 -4 Eample 2.7. Graph the region represented b the inequalit Solution: 2 + > 2 4 2 + = -2 2 + > -2 2-6 -4-2 2 4 6 8-2 -4 To find the region represented b a sstem of inequalities ou need to find the set of points represented b each inequalit in turn and then find the region which is the intersection of all of these sets of points.
52 Chapter 2: Geometric Linear Programming Eample 2.8. Find the region for the sstem of inequalities 2 3 6 2 + > 2 Also find all points of intersection of the straight lines bounding the region. Solution: 2 + = -2 4 2 2 3 = 6-6 -4-2 2 4 6 8-2 (0,-2) -4 Eample 2.9. Find the region represented b the sstem of inequalities + 2 4 3 6 < 3 Also find all points of intersection of the straight lines bounding the region. Solution: 4 3 = 6 2 (16/7,6/7) (3,1/2) -6-4 -2 2 4 6 8-2 + 2 = 4-4 = 3
Section 2.1: Graphing Inequalities 53 Eample 2.10. Find the region represented b the sstem of inequalities Solution: 2 + 3 6 + 1 2 < -2 + > 1-2 + 3 > 6 4 2-2 + 3 < 6 + > 1 > -2-6 -4-2 2 4 6 8 < -2 + < 1-2 + 3 < 6-2 -4 As can be seen from the graph, the region represented b these inequalities is empt since no region is the intersection of all 3 inequalities. One warning: ou ll have to be graphing regions like these in Section 2.3, when ou re solving linear programming problems. As was indicated earlier, ou need to make accurate graphs and ou ll need to label the boundar lines and, in particular, find and label the corner points for the region. To get used to doing that, make sure ou do it for each of the eercise problems that follow. Problems In Eercises 1-10, graph the inequalit. 1. 2 4 16 2. 3. 2 1 4. 5 + 7
54 Chapter 2: Geometric Linear Programming 5. 3 4 + 5 3 6 6. 9 + 2 > 18 7. 3 6 < 15 8. 3 5 5 4 < 0 9. 1.2 > 4.8 10 10. 3 15 + 5 In each of the following problems, graph the region determined b the sstem of inequalities. Find the coordinates of all corner points of the region and label them on the graph. If the region is empt, sa so. 11. 2 4 12 12. 3 1 + 3 2 + 4 8 2 13. 6 + 3 24 14. 1 3 + 6 < 30 5 < 15 1 + 10 5 15. 8 16. 5 + 2 25 2 + 22 4 3 3 4 2 0 0 0 17. + 16 18. 5 + 3 20 + < 8 + 3 16 < 8 4 + 3 40 + 4 0 > 4 0 0 19. 4 + 20 20. + 2 2 + 15 3 + 3 15 + 4 0 + 1 0 21. 4 + 3 24 + 7 3 + 4 24 0 0
Section 2.2: Setting Up Linear Programming Problems 55 2.2 Setting Up Linear Programming Problems As indicated in the introduction to this chapter, linear programming problems involve maimizing or minimizing a linear function (called the objective function) subject to a set of constraints which are epressed as linear inequalities. This section will be devoted to demonstrating how to set up linear programming problems. That means looking at a problem which is stated in everda language and figuring out how to translate it into mathematics. This step is usuall the most difficult one for students to handle and is the one where most of the serious mistakes occur. But, obviousl, ou can t hope to solve a problem correctl if ou don t have it set up right, so it s important to learn how to translate correctl. Let s look at the general procedure and then several eamples. There are some rules of thumb which can be applied in most cases and will probabl make our job easier. In setting up an linear programming problem, ou need to determine three things. The first one is: 1. What variables or unknowns are involved? Once ou know that, then ask ourself: 2. What quantit is to be maimized or minimized and how do I epress that quantit in terms of m unknowns? These two questions can generall be answered b reading our problem and finding the part of the problem that asks a question. The third step is: 3. What constraints do I have? How can I epress those constraints in terms of m unknowns? (In linear programming problems, this step results in a set of linear inequalities.) Eample 2.11. A potter is making cups and plates. It takes her 6 minutes to make a cup and 3 minutes to make a plate. Each cup uses 3/4 lb. of cla and each plate uses one lb. of cla. She has 20 hours available for making the cups and plates and has 250 lbs. of cla on hand. She makes a profit of $2 on each cup and $1.50 on each plate. How man cups and how man plates should she make in order to maimize her profit? Solution (setting up the problem): In this problem, the question is: How man cups and how man plates should she make in order to maimize her profit? So the variables are: = number of cups the potter makes = number of plates the potter makes That takes care of step 1 in the procedure. What about step 2? It s answered in the same phrase how man cups and how man plates should she make in order to maimize her profit? Clearl, profit is the quantit that is supposed to be maimized.
56 Chapter 2: Geometric Linear Programming But how do ou epress profit in terms of the unknowns? Well, since her profit is $2/cup and $1.50/plate, if the letter P represents her total profit, then she ll make a total profit of: so P = ($2/cup)( cups) + ($1.50/plate)( plates), P = $2 + $1.50 (Note that the cup units and the plate units cancel out so the P units are $ which, of course, makes sense for this problem.) That shows how to handle the first two steps in translating. How about the third step? Well there are some obvious constraints in this problem. First, the potter has a limit on the amount of time she has available for making the cups and plates (20 hours). Second, she has a finite amount of cla (250 lbs.). Those constraints need to be epressed in terms of the unknowns, and. If she has 20 hours available, and she makes cups and plates and each cup takes 6 minutes and each plate takes 3 minutes, then or (6 min./cup)( cups) + (3min./ plate)( plates) (20 hrs./min.)(60 min./hr.) 6 + 3 1200 (You use here since she s not required to use all her time. The units on both sides are minutes.) If each cup takes 3/4 lb. of cla and each plate takes 1 lb. of cla and she has 250 lbs. of cla available, then if she makes cups and plates, or (3/4 lb. of cla/cup)( cups) + (1 lb. of cla/plate)( plates) 250 lbs. of cla.75 + 250. (Again, she s not required to use all her cla. The units on both sides are lbs. of cla.) It s not uncommon to have a problem in which some of the constraints are obvious (if ou happen to think of them) but are unstated. These constraints generall take the form of assuming that the variables are nonnegative. That s certainl the case in this problem. Clearl, the potter can t make a negative number of cups or plates. However, even if the constraints are obvious, the need to be stated. As will be seen in the net section, when it comes to solving the problem, the constraints 0 0 are necessar to the solution of the problem. Thus, a statement of all the constraints in this problem is: 6 + 3 1200.75 + 250 0 0
Section 2.2: Setting Up Linear Programming Problems 57 A complete translation of the problem into a mathematical model then is: Let = number of cups the potter makes = number of plates the potter makes Maimize P = 2 + 1.5 subject to the constraints: 6 + 3 1200.75 + 250 0 0 Eample 2.12. A plant makes aluminum and copper wire. Each pound of aluminum wire requires 5 kwh of electricit and 1/4 hr. of labor. Each pound of copper wire requires 2 kwh of electricit and 1/2 hr. of labor. Production of copper wire is restricted b the fact that raw materials are available to produce at most 60 lbs./da. Electricit is limited to 500 kwh/da and labor to 40 person hrs./da. If the profit from aluminum wire is $.25/lb. and the profit from copper is $.40/lb., how much of each should be produced to maimize profit and what is the maimum profit? Solution (setting up the problem): Taking this one step at a time, the answer to step 1 is contained in the question part of this problem when ou re asked, How much of each (aluminum wire and copper wire) should be produced? So the unknowns are: = number of lbs. of aluminum wire = number of lbs. of copper wire Once the unknowns are determined, rereading the question indicates that it s the profit that is to be maimized. Clearl the profit can be epressed as: P = $.25 + $.40 Again, the rest of the problem contains the constraints. No more than 60 lbs. of copper can be produced each da, no more than 500 kwh of electricit are available each da and no more than 40 person hrs of labor are available each da. Using the rest of the information in the problem ields the following constraints: 1. lbs. of copper 60 lbs. of copper 2. (5 kwh/lb. of aluminum wire)( lbs. of aluminum wire) + (2 kwh/lb. of copper wire)( lbs. of copper wire) 500 kwh 3. (1/4 hr. of labor/lb. of aluminum wire)( lbs. of aluminum wire) + (1/2 hr. of labor/lb. of copper wire)( lbs. of copper wire) 40 hrs. of labor These inequalities reduce to: 60 5 + 2 500.25 +.5 40 It s also necessar to include, as before: 0, 0
58 Chapter 2: Geometric Linear Programming So the complete statement of the problem is: Let = number of lbs. of aluminum wire produced each da = number of lbs. of copper wire produced each da. Maimize P =.25 +.40 subject to the constraints: 60 5 + 2 500.25 +.5 40 0 0 Eample 2.13. A diet is to contain at least 600 units of vitamins, 600 units of minerals and 1800 calories. Two foods are available: A, which costs $.04/unit and B which costs $.06/unit. A unit of food A contains 2 units of vitamins, 1 unit of minerals and 4 calories; a unit of food B contains 1 unit of vitamins, 4 units of minerals and 4 calories. Find the minimum cost for a diet that consists of a miture of these two foods and also meets the minimum nutrition requirements. Solution (setting up the problem): Proceeding a step at a time, it s necessar to first determine the unknowns. The last sentence sas to find a minimum cost for a diet consisting of a miture of foods A and B, so it looks like the unknowns are: = number of units of food A = number of units of food B The cost of the diet is to be minimized so, letting C = cost of the diet, means minimizing the objective function: C = ($.04/unit of A) ( units of A) + ($.06/unit of B) ( units of B). Since the diet has to have at least 600 units of vitamins, 600 units of minerals and 1800 units of calories, the constraint inequalities are: (2 units of vitamins/unit of A) ( units of A) + (1 unit of vitamins/unit of B) ( units of B) 600 units of vitamins (1 unit of minerals/unit of A) ( units of A) + (4 units of minerals/unit of B) ( units of B) 600 units of minerals (4 calories/unit of A) ( units of A) + (4 calories/unit of B) ( units of B) 1800 calories Simplifing these inequalities and including the obvious ones which require that the variables be nonnegative gives: 2 + 600 + 4 600 4 + 4 1800 0
Section 2.2: Setting Up Linear Programming Problems 59 0 So the complete statement of the problem is: Let = number of units of A in the diet = number of units of B in the diet. Minimize C = $.04 + $.06 subject to the constraints: 2 + 600 + 4 600 4 + 4 1800 0 0 Eample 2.14. A manufacturer of outdoor furniture makes wooden couches, tables and chairs, each of which must be fabricated, assembled and finished. Each table requires 1 hour to fabricate, 3 hours to assemble and 4 hours to finish and returns a profit of $10. Each couch requires 6 hours to fabricate, 10 hours to assemble and 8 hours to finish and returns a profit of $12. Each chair requires 1 hour to fabricate, 1 hour to assemble and 1 hour to finish and returns a profit of $6. The manufacturer has 12,000 staff hours available each week for fabrication, 30,000 staff hours for assembling and 40,000 staff hours for finishing. How man units of each tpe of furniture should the manufacturer produce each week in order to maimize his profit? Solution (setting up the problem): Looking at the question asked in this problem, it s eas to see that the variables are: = number of tables = number of couches z = number of chairs The objective function (which is to be maimized) is represented b The constraint inequalities are: P = $10 + $12 + $6z. + 6 + z 12,000 3 + 10 + z 30,000 4 + 8 + z 40,000 0 0 z 0 Eamples 2.11, 2.12 and 2.13 will be solved in Section 3 but ou ll have to wait until Chapter 3, when the simple method is studied, in order to be able to solve Eample 2.14. The reason Eample 2.14 isn t solved using the methods of this chapter is that this eample involves more than 2 variables which means that the inequalities can t be graphed in two dimensional space (the plane). While it s possible to graph the
60 Chapter 2: Geometric Linear Programming inequalities in three dimensional space, the simple method of Chapter 3 provides a better alternative. Problems In each of the following eercises, set up the linear programming problem. In other words, identif the variables to be used, state the function to be maimized or minimized, and state all the constraint inequalities. Do not tr to solve the problem. 1. A farmer has a 320 acre farm on which she plants two crops: corn and sobeans. For each acre of corn planted, her epenses are $50 and for each acre of sobeans planted, her epenses are $100. Each acre of corn requires 100 bushels of storage and ields a profit of $60; each acre of sobeans requires 40 bushels of storage and ields a profit of $90. If the total amount of storage space available is 19,200 bushels and the farmer has onl $20,000 on hand, how man acres of each crop should she plant in order to maimize her profit? What will her profit be if she follows this strateg? 2. A dietitian wants to design a breakfast menu for certain hospital patients. The menu is to include two items X and Y. Suppose that each ounce of X provides 2 units of vitamin C and 2 units of iron and each ounce of Y provides 1 unit of vitamin C and 2 units of iron. Suppose the cost of X is 4 /ounce and the cost of Y is 3 /ounce. If the breakfast menu must provide at least 8 units of vitamin C and 10 units of iron, how man ounces of each item should be provided in order to meet the iron and vitamin C requirements for the least cost? What will this breakfast cost? 3. Sharon Brd wants to invest no more than $5000 in two stocks, A and B. Stock A is considered speculative and stock B is considered conservative. She wants to invest no more than $4000 in stock B and at least $600 in stock A. The average return is 10% for stock A (if the market doesn t take a nose dive) and 8% in stock B. To protect herself in case the market goes down sharpl, she decides her investment in stock B should be at least 1/3 her investment in stock A. How much should she invest in each stock to maimize her return on her investments? 4. A compan manufactures checker sets and chess sets. Suppose each da the compan has available 1900 boards (which can be used for both games) and 80,000 units of wood for making pieces. Each checker set uses 20 units of wood and each chess set uses 80 units of wood. The distributors the compan sells to can take up to 1250 checker sets per da and up to 750 chess sets per da. The compan makes a profit of $1 on each checker set and $1.25 on each chess set. How man checker sets and how man chess sets should the compan make each da in order to maimize its profits? What is the profit per da using this strateg?
Section 2.2: Setting Up Linear Programming Problems 61 5. Azalea State Creamer makes ice cream bars and ice cream sandwiches. It can make no more than a combined total of 4000 boes of bars and sandwiches each da but it has a market for up to 2400 boes of bars and up to 2000 boes of sandwiches. It can sell no more than three times as man bars as sandwiches. If the compan makes 6 on each bo of bars and 5 on each bo of sandwiches, find the number of boes of bars and sandwiches that should be made to maimize the profit. Find the profit per da using this strateg. 6. Michigan Polar Products makes downhill and cross-countr skis. A pair of downhill skis requires 2 man-hours for cutting, 1 man-hour for shaping and 3 man-hours for finishing while a pair of cross-countr skis requires 2 man-hours for cutting, 2 manhours for shaping and 1 man-hour for finishing. Each da the compan has available 140 man-hours for cutting, 120 man-hours for shaping and 150 man-hours for finishing. How man pairs of each tpe of ski should the compan manufacture each da in order to maimize profit if a pair of downhill skis ields a profit of $10 and a pair of cross-countr skis ields a profit of $8? 7. Tom decides to adopt a vegetarian diet consisting of fruits, grains and vegetables. His minimum dail requirements are 14 units of protein, 16 units of carbohdrates, and 12 units of fiber. Suppose a serving of fruits can suppl him with 1 unit of protein, 2 units of carbohdrates and 1 unit of fiber while a serving of grains provides 3 units of protein, 2 units of carbohdrates, and 3 units of fiber. A serving of vegetables provides 4 units of protein, 3 units of carbohdrates and 2 units of fiber. If fruit costs 30 per serving, grains cost 60 per serving and vegetables cost 70 per serving, how man servings of each tpe of food should he eat per da in order to satisf his dail food requirements at minimum cost? 8. And s Fanc Cookie Compan packages 3 cookie assortments, the Fanc assortment, the Delue assortment and the Supreme assortment. The Fanc assortment has 15 chocolate chip cookies and 3 chocolate chocolate chip cookies. The Delue assortment has 12 chocolate chip cookies, 3 chocolate cookies and 3 chocolate chocolate chip cookies while the Supreme assortment has 9 chocolate chip, 6 chocolate and 3 chocolate chocolate chip cookies. The compan makes 2100 chocolate chip cookies per minute, 1200 chocolate cookies per minute and 900 chocolate chocolate chip cookies per minute. The Fanc assortment sells for $1.50, the Delue for $1.75 and the Supreme for $2. How man of each tpe of assortment should be made per minute in order to maimize revenue? 9. O David s is tring to maimize profit for Big O s, Half Pounders and Turke O Nuggets. The have determined the make a profit of 60 on each Big O, 50 on each Half Pounder and $1 on each Turke O Nugget. Each product takes cooking time and assembl time. Big O s use 4 minutes of cooking time, Half Pounders use 3 minutes of cooking time and Turke O Nuggets use 5 minutes of cooking time. Big O s use 2 minutes of assembl time while Half Pounders and O Nuggets each use 1 minute of assembl time. There are 228 minutes of cooking time and 120 minutes of assembl time available at the lunch hour. The know from previous
62 Chapter 2: Geometric Linear Programming eperience that the must produce a minimum of 20 Big O s and 26 Half Pounders. How man of each tpe should the produce to maimize profit? 10. A famil building an earth-sheltered house needs at least 200 cubic ards of fill dirt and at least 50 cubic ards of top soil. Supplier A can provide at most 100 cubic ards of fill dirt and at most a combined total of 125 cubic ards of the two tpes of dirt. Supplier B can provide at most 30 cubic ards of top soil and at most a combined total of 150 cubic ards of the two tpes of dirt. The cost per cubic ard of each tpe of dirt from each supplier is given below. Find the amount of each tpe of dirt that should be provided b each supplier in order to minimize the cost. Fill dirt Topsoil A $4 per cu.d. $25 per cu. d. B $5 per cu.d. $20 per cu.d. 11. Pulverizer Gravel Compan has a quarr on Duraleigh Road and one on HW 1. The compan needs to provide gravel to 3 locations in Raleigh; one on Avent Ferr Road, one on Si Forks Road and one on Downtown Boulevard. The Duraleigh quarr can provide at most 600 tons/da and the one on HW 1 can provide at most 800 tons/da. The Avent Ferr Road, the Si Forks Road and the Downtown Boulevard locations require 400 tons, 400 tons and 500 tons of gravel dail respectivel. The costs for gravel delivered from the Duraleigh quarr to Avent Ferr Road, Si Forks Road and Downtown Boulevard are $8, $7 and $9 per ton respectivel while the costs for gravel delivered from the HW 1 quarr to the three locations are $11, $10 and $8 per ton respectivel. What trucking schedule will enable the compan to meet the demand for gravel and at the same time keep the cost of delivering the gravel to a minimum? (Hint: Let 1 = the number of tons of gravel delivered from the Duraleigh quarr to Avent Ferr Road. Then there are 5 other variables in this problem.) 12. A compan makes two tpes of sofas, regular and long, at two locations, one in Hickor and one in Lenoir. The plant in Hickor has a dail operating budget of $45,000 and can produce at most 300 sofas dail in an combination. It costs $150 to make a regular sofa and $200 to make a long sofa at the Hickor plant. The Lenoir plant has a dail operating budget of $36,000, can produce at most 250 sofas dail in an combination and makes a regular sofa for $135 and a long sofa for $180. The compan wants to limit production to a maimum of 250 regular sofas and 350 long sofas each da. If the compan makes a profit of $50 on each regular sofa and $70 on each long sofa, how man of each tpe should be made at each plant in order to maimize profit? What is the maimum profit?
Section 2.2: Setting Up Linear Programming Problems 63 13. A grocer store stocks 4 different sizes of tomato sauce; A, B, C, and D. The profit per case of each size is $1, $1.50, $1.50, and $2. The amounts of storage space per case of each size is 3, 6, 4, and 12 cubic feet respectivel. The amounts of time required for unpacking and shelving a case of each size are 2, 3, 4, and 4 minutes respectivel. From past eperience the store bus at least 2 times as much of size A as the other three sizes combined. Assume the store gets in a shipment weekl and has 4800 cubic feet of storage and 3600 minutes of unpacking and shelving time available per week. How man cases of each size of tomato sauce should be purchased each week in order to maimize profit? 2.3 Solving Problems Using the Geometric Method Now that ou know how to set up linear programming problems, let s look at a technique known as the Geometric Method (or Graphical Method or Method of Corners) which can be used to solve linear programming problems. To see wh it works, let s look at the first eample that was set up in Section 2.2. Eample 2.11 (continued): A potter is making cups and plates. It takes her 6 minutes to make a cup and 3 minutes to make a plate. Each cup uses 3/4 lb. of cla and each plate uses 1 lb. of cla. She has 20 hours available for making the cups and plates and has 250 lbs. of cla on hand. She makes a profit of $2 on each cup and $1.50 on each plate. How man cups and how man plates should she make in order to maimize her profit? Solution: This problem was stated in mathematical terms as: Let = number of cups the potter makes = number of plates the potter makes Maimize P = 2 + 1.5 subject to 6 + 3 1200.75 + 250 0 0 The graph of the region in the plane that s described b the set of constraint inequalities is:
64 Chapter 2: Geometric Linear Programming 400 300 200 100 6 + 3 = 1200 (0,250) (120,160).75 + = 250 (0,0) 100 (200,0) 300 400 This region (shaded in the graph) is known as the feasible region or feasible set, because the coordinates of the points in this region are the onl values of and that can be used in computing values for the profit P, since those values of and are the onl values that satisf all of the constraints. Now suppose P = 2 + 1.5 is graphed for various values of P. Then a set of lines that have the same slope (m = 4/3) but different intercepts (depending on the value used for P) will be graphed. If these parallel lines are superimposed on the graph above, the following picture is obtained: 400 300 (0,250) 200 (120,160) 100 (0,0) 100 (200,0) 300 400 P = 600 P = -400 P = -150 P = 0 P = 300 P = 200
Section 2.2: Solving Problems Using the Geometric Method 65 You can see that for some values of P (P = 400, P = 150, P = 600), the graph of the objective function P = 2 + 1.5 doesn t intersect the feasible region. The values of and that gave those values of P are not in the feasible region and thus don t satisf the constraints. In fact, the onl values of P that are allowed are the ones for which the graph of P = 2 + 1.5 does intersect the feasible region (e.g. P = 200 or P = 300). So what are those values of P? Well, if ou visualize sweeping the line for P = 2 + 1.5 from the bottom left part of the plane to the top right, ou can see that the line for P doesn t intersect the feasible region at all until the value for P increases to 0. As P increases from 0 through positive values of P, the line continues to intersect the region, but eventuall the value of P gets large enough that the line no longer intersects the feasible region. The last value of P (and the largest) for which the line still intersects the region is the value obtained for P at the corner point (120, 160). Now that s still an infinite number of values for P (from P evaluated at (0, 0) to P evaluated at (120, 160)) but since it s a maimum value for P that s needed, the problem becomes much simpler. There s onl one maimum value for P and that occurs at just one point the corner point (120, 160). So the maimum value of P on the feasible region is P = 2(120) + 1.5(160) = 480. It s also worth noting that the smallest value of P on the feasible region occurs at the corner point (0, 0). The ideas that have been used in this eample form the basis for the proof of a theorem. This theorem is the cornerstone of the Geometric method for solving linear programming problems. Before the theorem is stated, some terms which are used in the theorem need to be defined. Definition: Closed polhedral set A set in the plane is a closed polhedral set if the boundar of the set is formed b straight lines and the boundar lines are included in the set. Eample 2.15. Figure (a) is a closed polhedral set but neither Figure (b) nor Figure (c) is a closed polhedral set. (a) (b) not closed (c) not polhedral
66 Chapter 2: Geometric Linear Programming If ou look at the feasible sets associated with the eamples which were set up in Section 2.2, ou ll see that the re all closed polhedral sets. Theorem: Suppose C is a closed polhedral set in the plane and ƒ = a + b is a linear function. If ƒ has a maimum value on C the maimum value occurs at a corner point of C. If ƒ has a minimum value on C then that minimum value occurs at a corner point of C. If ƒ has the same value at two adjacent corner points of C, then ƒ has that value at ever point on the line segment joining those two corner points. The onl time ƒ ma not take on a maimum (minimum) value on a closed polhedral set is if the set is unbounded. If ou look at Eamples 2.11, 2.12 and 2.13 which were set up in Section 2.2 and graph the feasible sets associated with those problems, ou see that the regions in Eamples 2.11 and 2.12 are both bounded, so an linear function ƒ = a + b which is to be maimized (or minimized) on the feasible set will be maimized (or minimized) at a corner point. However, in the case of Eample 2.13, the region is unbounded. Eample 2.13 (continued): Minimize C = $.04 + $.06 subject to 2 + 600 + 4 600 4 + 4 1800 0, 0
Section 2.2: Solving Problems Using the Geometric Method 67 600 500 2 + = 600 400 300 (150,300) 200 100 + 4 = 600 4 + 4 = 1800 (400,50) 100 200 300 400 500 600 If the cost function C = $.04 + $.06 were to be maimized, it wouldn t be possible to do it since and can be made as big as ou want and the point would still be in the feasible region. However, C is to be minimized, which can be done, because sliding the line for C =.04 +.06 to the bottom left corresponds to looking at smaller values of C. In fact, the furthest toward the bottom left that the line for C can be moved and still intersect the feasible set is again a corner point of the feasible set. However, rather than tring to eeball which corner point gives the minimum, let s use the theorem. It sas that if C has a minimum on the set, it has to occur at a corner point. Since it can be seen from the graph that C should have a minimum on the region, let s evaluate C =.04 +.06 at each of the corner points of the feasible region and pick the smallest value of C. Thus C(0, 600) =.04(0) +.06(600) = 36 C(150, 300) =.04(150) +.06(300) = 24 C(400,50) =.04(400) +.06(50) = 19 C(600, 0) =.04(600) +.06(0) = 24 An alternate wa of displaing this information is to make a table showing all the corner points and the cost associated with each. Corner point Cost =.04 +.06 (0,600) $36 (150,300) $24 (400,50) $19 (600,0) $24 In an case the minimum value of C is $19 which occurs when the diet consists of 400 units of food A and 50 units of food B.
68 Chapter 2: Geometric Linear Programming If the theorem is applied to Eample 2.11, which was eeballed, then P(0,0) = $0 P(120,160) = $480 P(200,0) = $400 P(0,250) = $375 Again, the answer is that the maimum of P is $480. It occurs when 120 cups and 160 plates are produced. The onl other thing that can happen is if the function ƒ in the theorem is maimized (minimized) at two adjacent corner points. That can happen if the slope of the graph of ƒ is the same as the slope of one of the boundar lines of the feasible set. (See the eample below.) Then an point on the line segment joining the two corner points will maimize (minimize) ƒ and ou can pick whichever point on that line segment ou want to use. Eample 2.16. Maimize P = 2 + subject to 6 + 3 1200.75 + 250 0 0 Solution: 400 300 200 6 + 3 = 1200 (0,250) (120,160) 100 (200,0).75 + = 250 (0,0) 100 200 300 400 Corner point P = 2 + (0,0) $0 (200,0) $400 (120,160) $400 (0,250) $250
Section 2.2: Solving Problems Using the Geometric Method 69 The maimum value assumed b P is $400, and this value is obtained at all points on the line segment joining (120,160) and (200,0), the two corner points where the maimum occurs. Procedure for solving linear programming problems using the geometric method: 1. Translate the problem into mathematical terms (as eplained in Section 2.2). 2. Graph the feasible set described b the constraint inequalities and find the coordinates of all the corner points. If the region is unbounded, determine whether itʼs possible for the objective function to obtain the desired etreme value. If not, write no solution. Otherwise go to step 3. 3. Evaluate the objective function at each of the corner points. 4. Find the corner point that makes the objective function a maimum (minimum). If thereʼs onl one such corner point then the value of the objective function at that point is the maimum (minimum). If there are two adjacent corner points that maimize (minimize) the objective function then the maimum (minimum) value of the objective function occurs at an point on the line segment joining the two corner points. Eample 2.12 (continued). It was necessar to maimize P =.25 +.40 subject to 60 5 + 2 500.25 +.5 40 0 0 Solution: Graphing the feasible set gives the following region:
70 Chapter 2: Geometric Linear Programming 200 5 + 2 = 500 100 (0,60) (40,60) = 60 (85,37.5).25 +.5 = 40 (0,0) (100,0) 200 The corner points are (0,0), (100,0), (85,37.5), (40,60), (0,60), and evaluating P at each of those points gives: Corner Point P =.25 +.40 (0,0) $0 (100,0) $25 (85,37.5) $36.25 (40,60) $34 (0,60) $24 So the maimum value of P is $36.25 and that value occurs when = 85 and = 37.5. Thus the maimum profit occurs when 85 lbs. of aluminum wire and 37.5 lbs. of copper wire are produced each da. Eample 2.17. A manufacturer of pickup trucks makes a standard version and a customized version. Each standard pickup requires 8 staff hours to assemble, 2 staff hours to paint, 2 staff hours to upholster and is sold for a profit of $90. Each customized pickup requires 18 staff hours to assemble, 2 staff hours to paint, 1 staff hour to upholster and is sold for a profit of $100. During each da 360 staff hours are available to assemble, 50 staff hours to paint and 40 staff hours to upholster the trucks. How man of each tpe of pickup should be produced each da in order to maimize the profit? Solution: Let = number of standard pickups produced each da = number of customized pickups produced each da The problem is to maimize P = 90 + 100 subject to
Section 2.2: Solving Problems Using the Geometric Method 71 8 + 18 360 2 + 2 50 2 + 40 0 0 The graph of the feasible set is: 40 2 + 2 = 50 30 2 + = 40 20 (9,16) 10 (15,10) 8 + 18 = 360 10 20 30 40 The corner points are (0,0), (20,0), (15,10), (9,16), (0,20). Evaluating P at each of the corner points gives: Corner Point Profit = 90 + 100 (0,0) 90(0) + 100(0) = 0 (20,0) 90(20) + 100(0) = 1800 (15,10) 90(15) + 100(10) = 2350 (9,16) 90(9) + 100(16) = 2410 (0,20) 90(0) + 100(20) = 2000 So the maimum value of P is $2410 which occurs when 9 standard pickups and 16 customized pickups are produced each da. Eample 2.18. A nutritionist working for a compan providing backpacking equipment is told to produce a package of food sufficient for one backpacker for three das. She must meet certain nutritional requirements and et keep the weight of the food at a minimum. She is considering a combination of two dehdrated foods which are packaged in plastic bags. Each bag of food A contains 4 units of protein, 2 units of fat, and 2 units of carbohdrates and weighs.3 pounds. Each bag of food B contains 3 units of protein, 1 unit of fat and 6 units of carbohdrates and weighs.2 pounds. A backpacker needs 42 units of protein, 18 units of fat and 30 units of carbohdrates for a three da backpacking trip. How man bags of each food should she put in the package?
72 Chapter 2: Geometric Linear Programming Solution: Let = number of bags of food A = number of bags of food B She wants to minimize W =.3 +.2 subject to: 4 + 3 42 2 + 18 2 + 6 30 0 0 The graph of the feasible set is: 20 4 + 3 = 42 15 2 + = 18 10 5 (6,6) (9,2) 2 + 6 = 30 5 10 15 20 The corner points are: (0,18), (6,6), (9,2), (15,0). Evaluating W at each of the corner points gives: W(0,18) =.3(0) +.2(18) = 3.6 W(6,6) =.3(6) +.2(6) = 3.0 W(9,2) =.3(9) +.2(2) = 3.1 W(15,0) =.3(15) + (0) = 4.5 So the minimum value of W is 3.0 lbs which occurs when 6 bags of food A and 6 bags of food B are put in the package.
Section 2.2: Solving Problems Using the Geometric Method 73 Problems 1. Maimize P = 3 + 2 2. Maimize P = 2 + subject to subject to + 5 + 2 30 3 3 + 30 0, 0 0, 0 3. Minimize C = 3 + 4. Maimize P = 2 + 3 subject to subject to + 2 16 2 + 3 30 3 + 2 24 5 2 10 0 0, 0 5. Find the minimum and maimum of 6. Minimize C = 10 + P = 2 + 3 subject to subject to + 2 4 + 2 12 + 8 4 + 16 3 + 12 0, 0 0, 0 7. Maimize P = 2 + 3 8. Maimize P = + 2 subject to subject to 1 2 + 6 + 3 3 + 8 2 0 + 5 0, 0 9. Find the minimum and maimum of P = 3 4 subject to 2 + 3 11 1 2 + 4 8 3 + 2 In problems 10-14, find the maimum and minimum values of the objective function subject to the constraints + 12 2 + 12 + 2 12 10. P = 3 + 11. P = + 4
74 Chapter 2: Geometric Linear Programming 12. P = + 13. P = 2 + 14. P = + 2 In problems 15-17, determine the maimum and minimum values of the objective function, if possible. If not possible, eplain wh not. 15. P = 3 2 16. P = + 2 subject to subject to + 3 6 + 1 3 + 6 3 2 0, 0 + 3 3 2 + 2 4 17. P = 2 3 subject to + 4 2 + 6 0, 0 Problems 18-23 are problems which ou have alread set up in Section 2.2. 18. A dietitian wants to design a breakfast menu for certain hospital patients. The menu is to include two items X and Y. Suppose that each ounce of X provides 2 units of vitamin C and 2 units of iron and each ounce of Y provides 1 unit of vitamin C and 2 units of iron. Suppose the cost of X is 4 /ounce and the cost of Y is 3 /ounce. If the breakfast menu must provide at least 8 units of vitamin C and 10 units of iron, how man ounces of each item should be provided in order to meet the iron and vitamin C requirements for the least cost? What will this breakfast cost? 19. A farmer has a 320 acre farm on which she plants two crops: corn and sobeans. For each acre of corn planted, her epenses are $50 and for each acre of sobeans planted, her epenses are $100. Each acre of corn requires 100 bushels of storage and ields a profit of $60; each acre of sobeans requires 40 bushels of storage and ields a profit of $90. If the total amount of storage space available is 19,200 bushels and the farmer has onl $20,000 on hand, how man acres of each crop should she plant in order to maimize her profit? What will her profit be if she follows this strateg? 20. A compan manufactures checker sets and chess sets. Suppose each da the compan has available 1900 boards (which can be used for both games) and 80,000 units of wood for making pieces. Each checker set uses 20 units of wood and each chess set uses 80 units of wood. The distributors the compan sells to can take up to 1250 checker sets per da and up to 750 chess sets per da. The compan makes a profit of $1 on each checker set and $1.25 on each chess set. How man checker
Section 2.2: Solving Problems Using the Geometric Method 75 sets and how man chess sets should the compan make each da in order to maimize its profits? What is the profit per da using this strateg? 21. Sharon Brd wants to invest no more than $5000 in two stocks, A and B. Stock A is considered speculative and stock B is considered conservative. She wants to invest no more than $4000 in stock B and at least $600 in stock A. The average return is 10% for stock A (if the market doesn't take a nose dive) and 8% in stock B. To protect herself in case the market goes down sharpl, she decides her investment in stock B should be at least 1/3 her investment in stock A. How much should she invest in each stock to maimize her return on her investments? 22. Michigan Polar Products makes downhill and cross-countr skis. A pair of downhill skis requires 2 man-hours for cutting, 1 man-hour for shaping and 3 man-hours for finishing while a pair of cross-countr skis requires 2 man-hours for cutting, 2 manhours for shaping and 1 man-hour for finishing. Each da the compan has available 140 man-hours for cutting, 120 man-hours for shaping and 150 man-hours for finishing. How man pairs of each tpe of ski should the compan manufacture each da in order to maimize profit if a pair of downhill skis ields a profit of $10 and a pair of cross-countr skis ields a profit of $8? 23. Azalea State Creamer makes ice cream bars and ice cream sandwiches. It can make no more than a combined total of 4000 boes of bars and sandwiches each da but it has a market for up to 2400 boes of bars and up to 2000 boes of sandwiches. It can sell no more than three times as man bars as sandwiches. If the compan makes 6 on each bo of bars and 5 on each bo of sandwiches, find the number of boes of bars and sandwiches that should be made to maimize the profit. Find the profit per da using this strateg. 24. A small business makes 3-speed and 10-speed biccles at two different factories. Factor A produces 16 3-speed and 20 10-speed bikes in one da while factor B produces 12 3-speed and 20 10-speed bikes dail. It costs $1000/da to operate factor A and $800/da to operate factor B. An order for 96 3-speed bikes and 140 10-speed bikes has just arrived. How man das should each factor be operated in order to fill this order at a minimum cost? What is the minimum cost? 25. A to compan has two part-time emploees, Sam and Carl, who make pogo sticks, skateboards, and uniccles. Sam can turn out 1 pogo stick, 3 skateboards and 3 uniccles per da. Carl can turn out 2 pogo sticks, 5 skateboards and 1 uniccle per da. The compan has an order for 10 pogo sticks, 28 skateboards, and 8 uniccles. Sam is paid $50 per da and Carl is paid $60 per da. How man das should Sam and Carl work to turn out the order as cheapl as possible? 26. A lake is to be stocked with trout and salmon. Two foods, A and B, are available in the lake. Each trout requires 2 units of A and 1 unit of B per da while each salmon requires 1 unit of A and 3 units of B per da. Onl 500 units of A and 900 units of B are available dail. If each trout weighs 2 lbs. and each salmon weighs 4 lbs., how