VIEWS: 2 PAGES: 21 POSTED ON: 10/3/2012 Public Domain
Maximum and Minimum Chapter 6 1 Find inequalities from region In the graph, find inequalities for regions A, B, C, D, E 2 4 x 3 y 24 Region A has 3 boundaries x 2 y 10 x0 So it is 4 x 3 y 24 x 2 y 10 x 0 3 x 2 y 10 4 x 3 y 24 Region B has 5 boundaries x y 4 x0 y0 So it is x 2 y 10 4 x 3 y 24 x y 4 x 0 y 0 4 x 2 y 10 4 x 3 y 24 Region C has 4 boundaries x y 4 x0 So it is x 2 y 10 4 x 3 y 24 x y 4 x 0 5 x 2 y 10 Region D has 3 boundaries 4 x 3 y 24 x y 4 So it is x 2 y 10 4 x 3 y 24 x y 4 6 x y 4 Region E has 3 boundaries 4 x 3 y 24 y0 So it is x y 4 4 x 3 y 24 y 0 7 Exercises 1. In the graph, find inequalities for regions F and G 8 Positive linear functions If a 0, b 0 then z ax by is called a positive linear function. If z c then the line ax by c is called a z-value line. 1) All z-value lines are parallel. 2) Value of z ax by is increasing from southwest to northeast For example z 2 x 3 y, draw 0-value, 6-value and 6 -value lines. 9 z-value lins of z 2 x 3 y 10 Maximum and Minimum In a finite polygon region, any positive linear function z ax by will reach its maximum and minimum values at the corner points. Maximum value point Minimum value point 11 No maximum in unbounded region In a unbounded region (of Quadrant I), a positive linear function z ax by only has the minimum value, but NO MAXIMUM No Maximum value Minimum value point 12 Example Find maximum value of the subject function z 3x 4 y in the region 2x y 4 x 2 y 4 x0 y0 First we draw 4 lines 2x y 4 x 2 y 4 x0 y0 Then we shadow the region 13 Next we have to decide the corner points. Among 4 corner points, 3 of them are already known. They are (0,0), (0,2) and (2,0). We need to solve the last one, which is the common point of lines 2 x y 4 and x 2 y 4 Therefore , we setup 2 x y 4 (1) x 2 y 4 (2) (1) 2(2) to get 5 y 12 12 y (3) 5 Plug into (2) 12 4 x 2 4 x 5 5 Corner points are (0,0), (0,2), (2, 0) and 4 12 , 5 5 14 Now we calculate value of z 3x 4 y at those 4 corner points and draw the following table Corner Value z = 3x + 4y points (0, 0) 0 (0, 2) 30+42 = 8 (2, 0) 32+40 = 6 4 12 , 5 5 3 4 5 4 12 5 12 Maximum The maximum value of z is 12 at point 4 12 , 5 5 15 Example Find maximum and minimum values of z x 10 y in the region x 4 y 12 x 2y 0 2y x 6 x6 First we draw 4 lines x 4 y 12 x 2y 0 2y x 6 x6 Then we shadow the region 16 Next we have to decide the corner points. Among 4 corner points, 3 of them are easy to be found. They are (0,3), (6,3) and (6,6). The last one is the common point of lines x 4 y 12 and x 2 y 0. Therefore we setup x 4 y 12 (1) x 2 y 0 (2) (1) (2), we can get 6 y 12 y 2 Plug into (2), x40 x 4 So the last coner point is (4, 2) 17 Now we calculate value of z x 10 y at those 4 corner points and draw the following table Corner Value z = x + 10y points (0, 3) 0+3(10)=30 (6, 3) 6+10(3) = 36 6+10(6) = 66 (6, 6) maximum (4, 2) 4+10(2) = 24 minimum The maximum value of z is 66 at point (6,6) The minimum value of z is 24 at point (4,2) 18 Exercises Find maximum and minimum values in the region for the functions 1. z 3 x 2 y 2. z 0.4 x 0.75 y (2, 8) (0, 12) (4, 6) (4, 8) (1, 4) (7, 3) (4, 1) (0, 0) (8, 0) 19 3. Find maximum and minimum of z 4 x y if they exist (0, 10) (2, 4) (5, 2) (15, 0) 20 4. Find maximum and minimum of z 3x 6 y in region x y 10 5 x 2 y 20 x 2y 4 21