Example 3.2 Graphical Solution Method

| 3.1a | a3.3 Background Information n The Monet Company produces two type of picture frames, which we label 1, 2. n The two types of frames differ with respect to size, shape and materials used. n Each type 1 frame requires 2 labor hours, and each type 2 frame requires 1hour of labor. n Each type 1 frame requires I ounce of metal, and each type 2 frame requires 2 ounces of metal.

| 3.1a | a3.3 Background Information -- continued n During the coming week Monet has access to 4000 hours of skilled labor, and 5000 ounces of metal. n Type 1 frame contributes a profit of $2.25 per frame, and type 2 contributes a profit of $2.60 per frame. n THE COMPANY WANTS TO MAXIMIZE ITS PROFITS.

| 3.1a | a3.3 Traditional Algebraic Method n In the traditional algebraic solution method we first identify the decision variables. n In this problem they are the number of frames of type 1 and 2. We label these x 1 and x 2. n Next we write total profit and the constraints in terms of the x’s. n Finally, since only nonnegative amounts can be produced, we add explicit constraints to ensure that the x’s are nonnegative.

| 3.1a | a3.3 Background Information n To illustrate the graphical approach, we will use a slightly different scaled down version of Monet’s product mix problem. n Now there are only two frame types,1 and 2, and only two scarce resources, labor hours and metal. n The algebraic model is given below: max 2.25x x 2 (profit objective) subject to 2x 1 + x 2  4000 (labor constraint) x1 + 2x2  5000 (metal constraint) x1, x2  0 (nonnegativity constraint)

| 3.1a | a3.3 Background Information -- continued n The objective implies that each type 1 frame contributes a profit of $2.25, whereas each type 2 frame contributes a profit of $2.60. n The first constraint is a labor hour constraint. There are 4000 hours available. Each type 1 frame required 2 labor hours, and each type 2 frame requires 1 labor hour. n Similarly, the second constraint is a metal constraint. There are 5000 ounces of metal available. Each type 1 frame requires 1 ounce of metal and each type 2 frame requires 2 ounces of metal. n Find the optimal product mix graphically.

| 3.1a | a3.3 Solution n The idea is to graph the constraints on a two- dimensional graph to see which points (x 1, x 2 ) satisfy all of the constraints. This set of points is labeled the feasible region. Then we see which point in the feasible region provides the largest profit. n The graphical solution appears on the next slide.

| 3.1a | a3.3 Graphical Solution

| 3.1a | a3.3 Solution -- continued n To produce the graph, we first locate the lines where the constraints hold as equalities. n For example, the line for labor is 2x 1 + x 2 = The easiest way to graph this is to find the two points where it crosses the axes. n Joining the points (0,4000) and (2000,0), we get the line where the labor constraint is satisfied exactly, that is, as an equality. n All points below and to the left of this line are also feasible; there are these are the points where less than the maximum number of 4000 labor hours are used.

| 3.1a | a3.3 Solution -- continued n We indicate the feasible side of the line by the short arrows pointing down to the left from the labor constraint line. n Similarly the metal constraint line crosses the axes at the points (0,2500) and (5000,0), so we join these two points to find the line where all 5000 ounces of metal are used. n Finally the points on or below both of these lines constitute the feasible region. These are the point below the heavy lines.

| 3.1a | a3.3 Solution -- continued n You can think of the feasible region as all points on or inside the figure formed by four points: (0,0), (0,2500), (2000,0), and the point where labor hour and metal constraint lines intersect. n The next step is to bring profit into the picture. We do this by constructing “isoprofit” lines – that is lines where total profit is a constant. Any such line can be written as 2.25x x 2 = P where P is a constant profit level. Solving for x2, we can put this equation in slope-intercept form: x 2 = P/2.60 – (2.25/2.60)x 1

| 3.1a | a3.3 Solution -- continued n This shows that any isoprofit line has slope – 2.25/2.60, and it crosses the vertical axis at the value P/2.60. Three of these isoprofit lines appear in the chart as dotted lines. n Therefore, to maximize profit, we want to move the dotted line up and to the right until it just barely touches the feasible region. n Graphically, we can see that the last feasible point it will touch is the point indicated in the figure, where the labor hour and metal constraint lines cross.

| 3.1a | a3.3 Solution -- continued n We can then solve two equations in two unknowns to find the coordinates of this point. They are x 1 = 1000 and x 2 = 2000, with a corresponding profit of P = $7450. n Note that if the slope of the isoprofit lines were much steeper, the the optimal point would be (2000,0). On the other hand,m if the slope were mush less steep, the optimal point would be (0,2500).These statements make intuitive sense. n If the isoprofit lines are steep, this is because the unit profit from frame type 1 is large relative to the unit profit from frame type 2.

| 3.1a | a3.3 Solution -- continued n The crucial point, however, is that only three points can be optimal: (2000,0), (0, 2500), or (1000, 2000), the three “corner” points (other than (0,0)) in the feasible region. n The best of these depends on the relative slopes of the constraint lines and isoprofit lines in the graph.