Solve the following L.P. problem graphically. Maximize Z = x + y subject to constraints. x + y 3 , 2x + 3y 18 , x 6 , x 0 , y 0 .

Solution to the Linear Programming Problem

Step 1: Identify the Objective Function and Constraints

The objective is to maximize $Z=x+y$ subject to the following constraints:

• $x+y\ge 3$
• $2x+3y\le 18$
• $x\le 6$
• $x\ge 0$
• $y\ge 0$

Step 2: Rewrite the Inequalities for Graphical Representation

To plot the constraints, rewrite the inequalities as equations:

1. $x+y=3$
2. $2x+3y=18$
3. $x=6$
4. $x=0$ (y-axis)
5. $y=0$ (x-axis)

Step 3: Plot the Constraints

1. $x+y=3$:

   • When $x=0$, $y=3$.
   • When $y=0$, $x=3$.
   • Draw a solid line through (0, 3) and (3, 0) and shade the region above the line (since $\ge$).

2. $2x+3y=18$:

   • When $x=0$, $y=6$.
   • When $y=0$, $x=9$.
   • Draw a solid line through (0, 6) and (9, 0) and shade the region below the line (since $\le$).

3. $x=6$:

   • Draw a vertical solid line at $x=6$ and shade the region left of the line (since $\le$).

4. $x\ge 0$ and $y\ge 0$:

   • Only the first quadrant is considered.

Step 4: Identify the Feasible Region

The feasible region is the area where all constraints overlap. The vertices of this region are found at the intersections of the boundary lines:

• Intersection of $x+y=3$ and $2x+3y=18$: Solve simultaneously: $$\{\begin{array}{l}x+y=3\\ 2x+3y=18\end{array}$$ From the first equation, $y=3-x$. Substitute into the second: $$2x+3(3-x)=18⟹2x+9-3x=18⟹-x=9⟹x=-9$$ This is not feasible (since $x\ge 0$). So, check other intersections.

• Intersection of $x+y=3$ and $x=6$: $$6+y=3⟹y=-3\text{(Not feasible)}$$

• Intersection of $2x+3y=18$ and $x=6$: $$2(6)+3y=18⟹12+3y=18⟹3y=6⟹y=2$$ So, the point is $(6,2)$.

• Intersection of $x+y=3$ and $y=0$: $$x=3⟹(3,0)$$

• Intersection of $2x+3y=18$ and $y=0$: $$2x=18⟹x=9\text{(But x≤6, so not feasible)}$$

• Intersection of $x+y=3$ and $x=0$: $$y=3⟹(0,3)$$

• Intersection of $2x+3y=18$ and $x=0$: $$3y=18⟹y=6⟹(0,6)$$

The feasible vertices are:

1. $(0,3)$
2. $(0,6)$
3. $(6,2)$

Step 5: Evaluate the Objective Function at Each Vertex

Calculate $Z=x+y$ at each feasible vertex:

• At $(0,3)$: $Z=0+3=3$
• At $(0,6)$: $Z=0+6=6$
• At $(6,2)$: $Z=6+2=8$

Step 6: Determine the Maximum Value

The maximum value of $Z$ is 8, achieved at the point $(6,2)$.