Question

Which of the following is a point in the feasible region determined by the linear inequations \(3 \mathrm{x}+2 \mathrm{y}\) \(\geq 6\) and \(8 x+7 y \leq 56\) ? (1) \((3,1)\) (2) \((-3,1)\) (3) \((1,-3)\) (4) \((-3,-1)\)

Short answer

(1) (3,1) (2) (-3,1) (3) (1,-3) (4) (-3,-1) Solution: By testing each point, we found that (1) (3,1) is the only point that lies in the feasible region determined by the given linear inequalities. Therefore, the correct answer is (1) (3,1).

Step-by-step solution

Understand the given inequalities

We are given the following linear inequalities: 1. \(3x + 2y \geq 6\) 2. \(8x + 7y \leq 56\) We need to find which of the given points satisfies both of these inequalities.

Test point (1) \((3,1)\)

Substitute \(x=3\) and \(y=1\) into the inequalities: 1. \(3(3) + 2(1) \geq 6 \Rightarrow 9 + 2 \geq 6 \Rightarrow 11 \geq 6\) (True) 2. \(8(3) + 7(1) \leq 56 \Rightarrow 24 + 7 \leq 56 \Rightarrow 31 \leq 56\) (True) Since both inequalities hold true, point \((3,1)\) lies in the feasible region.

Test point (2) \((-3,1)\)

Substitute \(x=-3\) and \(y=1\) into the inequalities: 1. \(3(-3) + 2(1) \geq 6 \Rightarrow -9 + 2 \geq 6 \Rightarrow -7 \geq 6\) (False) 2. \(8(-3) + 7(1) \leq 56 \Rightarrow -24 + 7 \leq 56 \Rightarrow -17 \leq 56\) (True) Since the first inequality is false, point \((-3,1)\) does not lie in the feasible region.

Test point (3) \((1,-3)\)

Substitute \(x=1\) and \(y=-3\) into the inequalities: 1. \(3(1) + 2(-3) \geq 6 \Rightarrow 3 - 6 \geq 6 \Rightarrow -3 \geq 6\) (False) 2. \(8(1) + 7(-3) \leq 56 \Rightarrow 8 - 21 \leq 56 \Rightarrow -13 \leq 56\) (True) Since the first inequality is false, point \((1,-3)\) does not lie in the feasible region.

Test point (4) \((-3,-1)\)

Substitute \(x=-3\) and \(y=-1\) into the inequalities: 1. \(3(-3) + 2(-1) \geq 6 \Rightarrow -9 - 2 \geq 6 \Rightarrow -11 \geq 6\) (False) 2. \(8(-3) + 7(-1) \leq 56 \Rightarrow -24 - 7 \leq 56 \Rightarrow -31 \leq 56\) (True) Since the first inequality is false, point \((-3,-1)\) does not lie in the feasible region.

Conclusion

After testing the given points, point \((3,1)\) is the only point that lies in the feasible region determined by the given linear inequalities. Therefore, the correct answer is (1) \((3,1)\).

Key concepts

Feasible Region

In the context of linear inequalities, the feasible region is a crucial concept to understand. Imagine a coordinate plane divided by lines representing inequalities. The feasible region is the portion of the plane where all conditions or constraints of the inequalities overlap or intersect.

If you plot the given inequalities, such as the ones in the exercise,

• \(3x + 2y \geq 6\)
• \(8x + 7y \leq 56\)

The feasible region is the area where the shaded sections from both inequalities meet.
This area contains points that satisfy all the inequality constraints simultaneously. Only points within or on the boundary of this region fulfill the requirements, making them potential solutions.

Graphical Method in Inequalities

The graphical method is a helpful visual tool for solving systems of inequalities. It involves plotting each inequality on a coordinate plane. This method helps identify the feasible region where the solutions to the inequalities lie. To use the graphical method effectively:

• Start by converting each inequality into an equation by temporarily replacing the inequality sign with an equal sign.
• Draw the line of the equation. Use a solid line if the inequality is \(\geq\) or \(\leq\) to indicate the line is included in the feasible region. Use a dashed line for \(<\) or \(>\), indicating the boundary is not part of the solution.
• Shade the region that satisfies the inequality. For example, for inequality \(3x + 2y \geq 6\), shade above the line.
• Repeat these steps for each inequality and observe where all shaded areas overlap. This overlap is your feasible region.

By graphing these separately and identifying overlapping shaded areas, students can visually find solutions.

Solution of Inequalities

Finding a solution to inequalities involves determining specific values for variables that satisfy all given conditions. It requires substituting potential solutions back into the original inequalities to verify their validity. By substituting possible points into each inequality:

• Check if the inequality holds true when the point is substituted.
• For example, in the given exercise point \((3,1)\), substitute into inequalities:

• \(3(3) + 2(1) = 11 \geq 6\): True
• \(8(3) + 7(1) = 31 \leq 56\): True

Since both conditions are satisfied at point \((3,1)\), it is considered a valid solution within the feasible region. This method ensures students understand checking each possible point individually.

Inequality Constraints

Inequality constraints are limitations or conditions expressed in the form of inequalities. They dictate the range and set of possible solutions available.

• Constraints are set using inequality symbols like \(\geq\), \(>\), \(\leq\), and \(<\).
• For example, in the inequality \(3x + 2y \geq 6\), the constraint is that any valid solution must yield a value greater than or equal to 6.
• These boundaries form the framework that shapes the feasible region on a graph.
• Understanding these constraints helps visualize where solutions can lie within the feasible region, guiding the selection of test points.

Recognizing the role of inequality constraints aids in systematically approaching these problems, ensuring solutions adhere to all given condition.