The feasible region along with corner points for a linear programming problem are shown in the graph. Write all the constraints for the given linear programming problem.
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To determine the inequality from a line on a graph, pick a test point in the feasible region (not on the line) and substitute its coordinates into the equation of the line. If the inequality holds true for the test point, then that inequality defines the feasible region relative to that line. For example, for \( 2x + y = 70 \), the origin (0, 0) is in the feasible region, and \( 2(0) + 0 = 0 \le 70 \), confirming \( 2x + y \le 70 \).
The feasible region is defined by the intersection of several linear inequalities. We need to determine the equations of the boundary lines and the corresponding inequalities that define the shaded region.
Constraint from the line passing through (35, 0) and (30, 10):
The equation of the line passing through \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( \frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1} \).
Using points (35, 0) and (30, 10):
$$\frac{y - 0}{x - 35} = \frac{10 - 0}{30 - 35}$$
$$\frac{y}{x - 35} = \frac{10}{-5}$$
$$\frac{y}{x - 35} = -2$$
$$y = -2(x - 35)$$
$$y = -2x + 70$$
$$2x + y = 70$$
The feasible region is below this line, so the inequality is \( 2x + y \le 70 \).
Constraint from the line passing through (0, 30) and (15, 25):
Using points (0, 30) and (15, 25):
$$\frac{y - 30}{x - 0} = \frac{25 - 30}{15 - 0}$$
$$\frac{y - 30}{x} = \frac{-5}{15}$$
$$\frac{y - 30}{x} = -\frac{1}{3}$$
$$3(y - 30) = -x$$
$$3y - 90 = -x$$
$$x + 3y = 90$$
The feasible region is below this line, so the inequality is \( x + 3y \le 90 \).
Constraint from the horizontal line passing through (30, 10) and extending leftwards:
This line is \( y = 10 \).
The feasible region is above this line, so the inequality is \( y \ge 10 \).
Since the feasible region is in the first quadrant (including the axes), we also have the non-negativity constraints:
$$x \ge 0$$
$$y \ge 0$$
Therefore, the constraints for the given linear programming problem are:
$$2x + y \le 70$$
$$x + 3y \le 90$$
$$y \ge 10$$
$$x \ge 0$$
$$y \ge 0$$
