Exam practice

Boundary x + y = 3 through (0, 3) and (3, 0), solid. Test O: 0 ≥ 3, false: shade the side away from O.

Boundary x − y = 0, that is y = x, through (0, 0) and (2, 2), solid. O is on the line, so test (1, 0): 1 ≥ 0, true: shade the side containing (1, 0), below the line.

Boundary 2x + 3y = 6 through (0, 2) and (3, 0), solid. Test O: 0 ≤ 6, true: shade O’s side (as in Example 1g).

Boundary 3x + 4y = 12 through (0, 3) and (4, 0), solid. Test O: 0 ≥ 12, false: shade the side away from O.

Boundary x − 2y = 3: x = 0 gives y = −3/2, and y = 0 gives x = 3; (1, −1) is on it too. Solid line through (3, 0) and (1, −1). Test O: 0 ≥ 3, false: shade the side away from O, below the line.

Boundary 3x − 2y = 4: x = 0 gives y = −2; x = 2 gives y = 1. Solid line through (0, −2) and (2, 1). Test O: 0 ≥ 4, false: shade the side away from O.

x + 2y = 6 through (0, 3) and (6, 0); O gives 0 ≤ 6, true. x ≥ 0: right of the Y-axis. The feasible region is to the right of the Y-axis and below the line: it is open, running on downwards, with one corner P(0, 3).

2x − y = 4 through (2, 0) and (0, −4); O gives 0 ≥ 4, false, so the region is on the side away from O, to the right. y ≥ 0: above the X-axis. The feasible region is open, to the right of the line and above the X-axis, with its corner at P(2, 0).

x + y = 1 through (1, 0) and (0, 1); O gives 0 ≤ 1, true. x − y = 1 through (1, 0) and (0, −1); O gives 0 ≥ 1, false. The lines cross at P(1, 0). The feasible region is the open wedge below both lines, between them, with its point at P(1, 0).

2x − y ≤ 0 means y ≥ 2x: above the line y = 2x. Is x − y ≤ 2 needed at all?

3x + 4y = 12 through (0, 3) and (4, 0), O’s side. 2x − y = 0 through (0, 0) and (1, 2); test (0, 1): −1 ≤ 0, true, so the side above the line. x − y = 2 through (2, 0) and (0, −2), O’s side. x, y ≥ 0: the first quarter.
The feasible region is the triangle O(0, 0), A(12/11, 24/11), C(0, 3). A is where y = 2x meets 3x + 4y = 12: 11x = 12, x = 12/11, y = 24/11. The line x − y = 2 doesn’t touch the triangle, so that constraint changes nothing.

2x + 3y = 6 through (0, 2) and (3, 0), O’s side. 3x − y = 0 (y = 3x) through (0, 0) and (1, 3); test (0, 1): −1 ≤ 0, true, so the side above it.
The feasible region is the triangle O(0, 0), A(6/11, 18/11), B(0, 2). A: y = 3x in 2x + 3y = 6 gives 11x = 6, x = 6/11, y = 18/11.

x + y = 2 through (2, 0) and (0, 2), O’s side. 2x − 3y = 6 through (3, 0) and (0, −2), O’s side (0 ≤ 6). x ≥ 0: right of the Y-axis; y ≤ 2: below y = 2.
The feasible region is the triangle P(0, 2), Q(12/5, −2/5), R(0, −2). Q: from x = 2 − y, 4 − 2y − 3y = 6, so y = −2/5, x = 12/5. The line y = 2 only touches it at P.

A(0, 3) and B(2, 0) give the intercepts 2 and 3:
$$\frac{x}{2} + \frac{y}{3} = 1 \;\Rightarrow\; 3x + 2y = 6$$
C(4, −3) is on it too: 12 − 6 = 6. O is in the region and 0 ≤ 6; the line is solid. 3x + 2y ≤ 6.

One line passes through O, so test with a point such as (2, −2), which is inside the region.

Line OC through (0, 0) and (3, 3): y = x, or x − y = 0.
Line AB through (−1, 1) and (2, 0): slope (0 − 1) ÷ (2 + 1) = −1/3, so
$$y = -\frac{1}{3}(x - 2) \;\Rightarrow\; x + 3y = 2$$
Test (2, −2), inside: 2 + 2 = 4 ≥ 0 and 2 − 6 = −4 ≤ 2.
x − y ≥ 0 and x + 3y ≤ 2.

Line AC through (0, 1) and (−1, −1): slope 2, y = 2x + 1, that is 2x − y = −1.
Line BC through (2, 0) and (−1, −1): slope (−1 − 0) ÷ (−1 − 2) = 1/3, so
$$y = \frac{1}{3}(x - 2) \;\Rightarrow\; x - 3y = 2$$
O is inside: 0 ≥ −1, 0 ≤ 2.
2x − y ≥ −1 and x − 3y ≤ 2.

The lines are those of Example 3: AB is 2x − y = −2 and CD is 3x + 4y = 12, with O on their allowed sides. This time the X-axis closes the region off below, so there is a third inequality: O lies on the X-axis, so test a point inside such as (1, 1), which gives 1 ≥ 0: y ≥ 0. 2x − y ≥ −2, 3x + 4y ≤ 12, y ≥ 0.

x + y = 6 through (6, 0), (0, 6); x − y = 4 through (4, 0), (0, −4); O is on the allowed side of both. They meet where 2x = 10: (5, 1).

Z: O 0, A(4, 0) 12, B(5, 1) 17, C(0, 6) 12.
Maximum 17 at (5, 1); minimum 0 at (0, 0).

x + y = 6: O’s side. x − y = 2 through (2, 0) and (0, −2): O gives 0 ≥ 2, false, so the side away from O. They meet at (4, 2).

P: A(2, 0) 6, B(6, 0) 18, C(4, 2) 12 + 10 = 22.
Maximum 22 at (4, 2); minimum 6 at (2, 0).

x − 2y = 1 through (1, 0) and (3, 1); x + y = 4 through (4, 0) and (0, 4); O is on the allowed side of both. They meet where 3y = 3: (3, 1).

Z: O 0, A(1, 0) 3, B(3, 1) 9 + 5 = 14, C(0, 4) 20.
Maximum 20 at (0, 4); minimum 0 at (0, 0). (B gives only 14: with y worth 5 and x only 3, the corner with the most y wins.)

x − 2y = 2 through (2, 0) and (4, 1); x + 2y = 8 through (8, 0) and (0, 4); O is on both allowed sides. Adding the equations: 2x = 10, x = 5, y = 3/2.

F: O 0, A(2, 0) 6, B(5, 3/2) 15 + 3 = 18, C(0, 4) 8.
Maximum 18 at (5, 3/2); minimum 0 at (0, 0).

There is no x ≥ 0 here, so the region reaches left of the Y-axis.

x + 2y = 10 through (10, 0) and (0, 5), O’s side; x + y = 1 through (1, 0) and (0, 1), the side away from O; between y = 0 and y = 4.
Corners: x + y = 1 meets y = 0 at (1, 0) and y = 4 at (−3, 4); x + 2y = 10 meets y = 0 at (10, 0) and y = 4 at (2, 4).

F: A(1, 0) 2, B(10, 0) 20, C(2, 4) 8, D(−3, 4) −2.
Maximum 20 at (10, 0); minimum −2 at (−3, 4).

Line AB through (3, 0) and (5, 3): slope 3/2, so
$$y = \frac{3}{2}(x - 3) \;\Rightarrow\; 3x - 2y = 9$$
O: 0 ≤ 9.
Line BC through (5, 3) and (0, 6): slope −3/5, so
$$y = 6 - \frac{3}{5}x \;\Rightarrow\; 3x + 5y = 30$$
O: 0 ≤ 30.
The inequalities are 3x − 2y ≤ 9, 3x + 5y ≤ 30, x ≥ 0, y ≥ 0.
F: O(0, 0) 0, A(3, 0) 60, B(5, 3) 100 + 90 = 190, C(0, 6) 180.
Maximum 190 at (5, 3); minimum 0 at (0, 0).