5.2 Q-2

Question Statement

Graph the feasible region of the following system of linear inequalities and find the corner points in each case:

(i)

$$2x + y \geq 10, \quad x + 4y \geq 12, \quad x + 2y \leq 10, \quad x \geq 0, \quad y \geq 0$$

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Background and Explanation

This question involves graphing the feasible region defined by a system of linear inequalities. Each inequality represents a half-plane, and the solution to the system is the region where all these half-planes overlap. The corner points of this feasible region can be found by solving the system algebraically and graphing the constraints. Understanding how to solve linear inequalities and finding the intersection points of lines is essential.

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Solution

To graph the feasible region, we start by finding the points of intersection of the boundary lines, which are the equations obtained from converting inequalities to equalities.

(i) Graphing the System

1. Graph the inequality $2x + y \geq 10$

   • Convert to equality: $2x + y = 10$
   • Set $x = 0$: $y = 10 \Rightarrow (0, 10)$
   • Set $y = 0$: $2x = 10 \Rightarrow x = 5 \Rightarrow (5, 0)$
   • Plot the line and shade above it because $y \geq 10 - 2x$.

2. Graph the inequality $x + 4y \geq 12$

   • Convert to equality: $x + 4y = 12$
   • Set $x = 0$: $4y = 12 \Rightarrow y = 3 \Rightarrow (0, 3)$
   • Set $y = 0$: $x = 12 \Rightarrow (12, 0)$
   • Plot the line and shade above it because $y \geq \frac{12 - x}{4}$.

3. Graph the inequality $x + 2y \leq 10$

   • Convert to equality: $x + 2y = 10$
   • Set $x = 0$: $2y = 10 \Rightarrow y = 5 \Rightarrow (0, 5)$
   • Set $y = 0$: $x = 10 \Rightarrow (10, 0)$
   • Plot the line and shade below it because $y \leq \frac{10 - x}{2}$.

4. Graph the inequalities $x \geq 0$ and $y \geq 0$ to restrict the solution to the first quadrant.

After plotting all these inequalities, the feasible region is the shaded region where all the inequalities overlap in the first quadrant.

Finding Corner Points

1. Intersection of $x + 4y = 12$ and $2x + y = 10$:
   • Solve the system:

$$x + 4y = 12 \quad \text{(eq. 1)}$$ $$2x + y = 10 \quad \text{(eq. 2)}$$

  Solve eq. 1 for $x$:

$$x = 12 - 4y$$

  Substitute into eq. 2:

$$2(12 - 4y) + y = 10$$ $$24 - 8y + y = 10$$ $$-7y = -14 \quad \Rightarrow y = 2$$

  Substitute $y = 2$ into $x = 12 - 4y$:

$$x = 12 - 4(2) = 4$$

  Hence, the corner point is $(4, 2)$.

2. Intersection of $x + 4y = 12$ and $x + 2y = 10$: - Set the system equal:

$$x + 4y = 12 \quad \text{(eq. 1)}$$ $$x + 2y = 10 \quad \text{(eq. 2)}$$

  Subtract eq. 2 from eq. 1:

$$(x + 4y) - (x + 2y) = 12 - 10$$ $$2y = 2 \quad \Rightarrow y = 1$$

  Substitute $y = 1$ into $x + 2y = 10$:

$$x + 2(1) = 10 \quad \Rightarrow x = 8$$

  Hence, another corner point is $(8, 1)$.

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Key Formulas or Methods Used

• Solving systems of linear equations: To find the intersection points of the boundary lines.
• Graphing linear inequalities: Graphing each inequality as an equation, shading the region that satisfies the inequality.

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Summary of Steps

1. Convert each inequality into an equation.
2. Graph the boundary lines for each equation and shade the appropriate region.
3. Find the intersection points by solving the system of equations.
4. The corner points of the feasible region are where the lines intersect.
5. Check the feasibility of each point in the context of all inequalities.