Question

Solve the following Linear Programming Problem graphically:
Maximize $Z = \frac{2x}{5} + \frac{3y}{10}$
subject to constraints:
$2x + y \le 1000$
$x + y \le 800$
$x, y \ge 0$.

Hint:

When evaluating fractions in objective functions, avoid calculating decimal equivalents immediately. Convert the entire function to a common denominator first, like $\frac{4x+3y}{10}$. This makes substituting whole coordinates much quicker and reduces basic calculation errors!

Answer 1

Concept: To solve a Linear Programming Problem (LPP) graphically, we execute the following systematic process:
• Treat inequalities as equations to determine boundary lines.
• Identify the half-planes determined by inequalities to locate the feasible region.
• Determine the corner points (vertices) of the bounded feasible region by finding intersection points.
• Apply the Corner Point Theorem by evaluating the objective function $Z$ at each vertex to find the maximum value.

Step 1: Finding points for the line $2x + y = 1000$
To plot the boundary line corresponding to the first constraint $2x + y \le 1000$:
• Set $x = 0$: $2(0) + y = 1000 \implies y = 1000$. This gives the y-intercept point $A(0, 1000)$.
• Set $y = 0$: $2x + 0 = 1000 \implies x = 500$. This gives the x-intercept point $B(500, 0)$. Testing the origin $(0,0)$ in the inequality $2(0) + 0 \le 1000 \implies 0 \le 1000$, which is true. Hence, the region containing the origin is part of the solution space for this constraint.

Step 2: Finding points for the line $x + y = 800$
To plot the boundary line corresponding to the second constraint $x + y \le 800$:
• Set $x = 0$: $0 + y = 800 \implies y = 800$. This gives the y-intercept point $C(0, 800)$.
• Set $y = 0$: $x + 0 = 800 \implies x = 800$. This gives the x-intercept point $D(800, 0)$. Testing the origin $(0,0)$ in the inequality $0 + 0 \le 800 \implies 0 \le 800$, which is true. Hence, the region containing the origin is also part of the solution space for this constraint.

Step 3: Finding the intersection point of the two boundary lines
Let the intersection point of the two lines be denoted by $E$. We solve the simultaneous linear system: \[ 2x + y = 1000 \quad \cdots (\text{Eq. 1}) \] \[ x + y = 800 \quad \cdots (\text{Eq. 2}) \] Subtracting Eq. 2 from Eq. 1: \[ (2x + y) - (x + y) = 1000 - 800 \] \[ x = 200 \] Substituting $x = 200$ back into Eq. 2: \[ 200 + y = 800 \implies y = 600 \] Thus, the intersection point of the two lines is $E(200, 600)$.

Step 4: Identifying the corner points of the bounded feasible region
The non-negativity constraints $x \ge 0, y \ge 0$ restrict our feasible region entirely to the first quadrant. By combining the regions bounded by both lines towards the origin, we find that the closed feasible region is a quadrilateral defined by the following corner points:
• $O(0, 0)$ — The origin
• $B(500, 0)$ — On the x-axis
• $E(200, 600)$ — The intersection point
• $C(0, 800)$ — On the y-axis

Step 5: Evaluating the Objective Function $Z$
To make calculations easier, let's rewrite the objective function $Z$ with a common denominator: \[ Z = \frac{2x}{5} + \frac{3y}{10} = \frac{4x + 3y}{10} \] Let's compute the value of $Z$ systematically at each of the four corner points:
• At point $O(0,0)$: \[ Z = \frac{4(0) + 3(0)}{10} = 0 \]
• At point $B(500, 0)$: \[ Z = \frac{4(500) + 3(0)}{10} = \frac{2000}{10} = 200 \]
• At point $E(200, 600)$: \[ Z = \frac{4(200) + 3(600)}{10} = \frac{800 + 1800}{10} = \frac{2600}{10} = 260 \]
• At point $C(0, 800)$: \[ Z = \frac{4(0) + 3(800)}{10} = \frac{2400}{10} = 240 \] Comparing all calculated values of $Z$ ($0, 200, 260, 240$), the maximum value obtained is $260$, which occurs precisely at the coordinates $(200, 600)$.