5.3 Q-3

Question Statement

Maximize the objective function:
$z = 2x + 3y$
Subject to the constraints:
$3x + 4y \leq 12 \quad \text{(1)}$
$2x + y \leq 4 \quad \text{(2)}$
$x \geq 0 \quad \text{(3)}$
$y \geq 0 \quad \text{(4)}$

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Background and Explanation

In this optimization problem, we are tasked with maximizing the function $z = 2x + 3y$, subject to a set of linear inequalities. To solve this, we will use linear programming techniques, specifically the graphical method. This involves plotting the constraints, identifying the feasible region, and calculating the value of the objective function at each vertex (corner point) of the feasible region.

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Solution

Step 1: Graph the Constraints

Constraint 1: $3x + 4y \leq 12$
Rearrange the inequality to find the boundary line:
$3x + 4y = 12$

Find the intercepts:

• Set $x = 0$, solve for $y$:
  $4y = 12 \quad \Rightarrow \quad y = 3$
  So, the point is $(0, 3)$.
• Set $y = 0$, solve for $x$:
  $3x = 12 \quad \Rightarrow \quad x = 4$
  So, the point is $(4, 0)$.

Constraint 2: $2x + y \leq 4$
Rearrange the inequality to find the boundary line:
$2x + y = 4$

Find the intercepts:

• Set $x = 0$, solve for $y$:
  $y = 4$
  So, the point is $(0, 4)$.
• Set $y = 0$, solve for $x$:
  $2x = 4 \quad \Rightarrow \quad x = 2$
  So, the point is $(2, 0)$.

Checking the feasibility:

• At $x = 0, y = 0$, both constraints hold:
  $0 + 0 < 12 \quad \text{and} \quad 0 + 0 < 4$
  Both are true, so the origin is part of the feasible region.

Step 2: Identify the Feasible Region

The feasible region is the intersection of the half-planes defined by the constraints, which form a quadrilateral in the first quadrant. We plot the lines $3x + 4y = 12$ and $2x + y = 4$ on the graph, and the feasible region is the area where all constraints are satisfied.

Step 3: Find the Corner Points

The corner points of the feasible region are the intersections of the boundary lines:

1. $(0, 3)$ from constraint 1.
2. $(4, 0)$ from constraint 1.
3. $(2, 0)$ from constraint 2.
4. The intersection of the two lines $3x + 4y = 12$ and $2x + y = 4$.

Finding the intersection of the lines:

Substitute the expression $y = 4 - 2x$ from constraint 2 into constraint 1: $3x + 4(4 - 2x) = 12$
Simplify: $3x + 16 - 8x = 12$
$-5x + 16 = 12$
$-5x = -4$
$x = \frac{4}{5}$

Substitute $x = \frac{4}{5}$ into $y = 4 - 2x$: $y = 4 - 2\left(\frac{4}{5}\right) = 4 - \frac{8}{5} = \frac{20}{5} - \frac{8}{5} = \frac{12}{5}$
Thus, the intersection point is $\left(\frac{4}{5}, \frac{12}{5}\right)$.

Step 4: Evaluate the Objective Function at the Corner Points

Now, we evaluate $z = 2x + 3y$ at each of the corner points:

• At $\left(\frac{4}{5}, \frac{12}{5}\right)$:
  $z = 2\left(\frac{4}{5}\right) + 3\left(\frac{12}{5}\right) = \frac{8}{5} + \frac{36}{5} = \frac{44}{5} = 8.8$

• At $(4, 0)$:
  $z = 2(4) + 3(0) = 8 + 0 = 8$

• At $(0, 3)$:
  $z = 2(0) + 3(3) = 0 + 9 = 9$

• At $(2, 0)$:
  $z = 2(2) + 3(0) = 4 + 0 = 4$

Step 5: Identify the Maximum Value

The maximum value of $z$ is 9, which occurs at the corner point $(0, 3)$.

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Key Formulas or Methods Used

• Objective function: $z = 2x + 3y$
• Linear inequalities: $3x + 4y \leq 12$, $2x + y \leq 4$
• Graphical method: Plot the constraints, find the feasible region, and evaluate the objective function at the corner points.

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Summary of Steps

1. Graph the constraints and identify the intercepts for each inequality.
2. Plot the feasible region formed by the constraints.
3. Identify the corner points by solving for the intersections of the boundary lines.
4. Evaluate the objective function $z = 2x + 3y$ at each corner point.
5. Determine the maximum value of $z$ and the corresponding corner point.

The maximum value of $z$ is 9 at the corner point $(0, 3)$.