Saturday, July 25, 2020

Systems of Linear Inequalities - Notes

Here are the notes for the Systems of Linear Inequalities Section in the Linear Programming Unit.  Let me know if there are any questions.

Systems of Linear Inequalities. Basic Procedure. Find the region of feasible solutions for: ax + by less than or equal to c and dx + ey less than or equal to f, where at least one of a and b are nonzero and at least one of d and e are nonzero. Step 1: Graph ax + by = c and dx + ey = f. Remember that less than or equal to and greater than or equal to uses a solid line and less than and greater than use a dashed line. Step 2: Find where the two lines, ax + by = c and dx + ey = f intersect. Step 3: Find the solutions to the inequalities ax + by less than or equal to c and dx + ey less than or equal to f, by substituting a point not on the lines and determining whether it is true or false. Shade the regions where it is true. Example 1: Graph the system: y less than or equal to 2x - 2 and y > -x + 3. Step 1: Graph y = 2x - 2 and y = - x + 3. For y = 2x - 2, for the x-intercept put y = 0 into y = 2x - 2 to get 0 = 2x - 2 implies that 2 = 2x implies that x = 1. So, the x-intercept is (1, 0).

y-intercept: Put x = 0 into y = 2x - 2 to get y = 2(0) - 2 = -2. So, (0, -2) is a point on y = 2x - 2. For y = -x + 3, for the x-intercept, put y = 0 into y = -x + 3 to get x = 3. So, (3, 0) is the x-intercept. For the y-intercept: Put = 0 into y = -x + 3 to get y = 3. So, (0, 3) is the y-intercept. Where do y = 2x - 2 and y = -x + 3 intersect? Solve the system: y = 2x - 2, y = -x + 3. So, 2x - 2 = -x + 3 implies that 3x - 2 = 3 implies that 3x = 5 implies that x = 5/3. Put x = 5/3 into y = - x + 3 to get y = 4/3. So, y = 2x - 2 an y = -x + 3 intersect at (5/3, 4/3). Now, put everything on the graph so far and label the regions 1 to 4. Region 1: Above y = -x + 3 and to the right of y = 2x - 2, Region 2: Above y = - x + 3 to the left of y = 2x - 2, Region 3: Below y = - x + 3 to the left of y = 2x - 2, Region 4: Belove y = - x + 3 and to the right of y = 2x - 2.

To see if Region 1 works, pick a point from there, say (5, 0) (try to pick a point where at least one of the coordinates is 0). Note that by substituting (5, 0) into the original system of equations we get that 0 is less than or equal to 2(5) - 2 = 8 and 0 > -5 + 3 = -2. Since both of those are true, we will shade Region 1. To see if Region 2 works, try (0, 5): Since 5 is less than or equal to 2(0) - 2 = -2 is false, the whole system is false. So, we do not shade Region 2. To see if Region 3 works, try (0,0). Since 0 less than or equal to 2(0) - 2 = -2 is false, the system is false. Thus, we do not shade Region 3. To see if Region 4 works, try (0, -5). The first equation holds: -5 is less than or equal to 2(0) - 2 but the second equation does not: -5 > -0 + 3 is false. Thus, we do not shade Region 4.

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