Maximize Z = 5x1 + 6x2 Subject to 2x1 + 3x2 ≤ 6, x1, x2 ≥ 0

Steps to Solve

1. Identify the constraints and the objective function

The objective function is given by:

$$ Z = 5x_1 + 6x_2 $$

The constraint is:

$$ 2x_1 + 3x_2 \leq 6 $$

Along with the non-negativity constraints:

$$ x_1, x_2 \geq 0 $$

2. Convert the inequality into an equation

To find the boundary line of the constraint, we can convert the inequality into an equation:

$$ 2x_1 + 3x_2 = 6 $$

3. Determine intercepts of the constraint line

To find the intercepts, set $x_1$ and $x_2$ to zero one at a time:

• For $x_1$-intercept:

Set $x_2 = 0$:

$$ 2x_1 + 3(0) = 6 \implies x_1 = 3 $$

This gives the point $(3, 0)$.

• For $x_2$-intercept:

Set $x_1 = 0$:

$$ 2(0) + 3x_2 = 6 \implies x_2 = 2 $$

This gives the point $(0, 2)$.

4. Plot the feasible region

Plot the points $(3, 0)$ and $(0, 2)$ on a graph, and shade the region that satisfies the constraint $2x_1 + 3x_2 \leq 6$ along with the non-negativity restrictions.

5. Graph the objective function

Graph the objective function by rearranging it into slope-intercept form:

$$ 6x_2 = Z - 5x_1 $$

For various values of $Z$, determine lines of the form $x_2 = \frac{Z - 5x_1}{6}$.

6. Find the optimal solution

Identify the corner points of the feasible region, which are:

• Point A: $(0, 0)$
• Point B: $(3, 0)$
• Point C: $(0, 2)$
• Point D: Intersection of $2x_1 + 3x_2 = 6$ and the axes (not applicable here since we already use intercepts).

Calculate the value of $Z$ at each corner point:

• At $(0, 0)$: $Z = 5(0) + 6(0) = 0$
• At $(3, 0)$: $Z = 5(3) + 6(0) = 15$
• At $(0, 2)$: $Z = 5(0) + 6(2) = 12$

7. Select the maximum value

The maximum value of $Z$ occurs at $(3, 0)$, where $Z = 15$.