Question

Solve the following Linear Programming Problem graphically :
Minimize $Z = 20x + 10y$
subject to the constraints:
$x + 2y \le 40$
$3x + y \ge 30$
$4x + 3y \ge 60$
$x, y \ge 0$

Hint:

Always substitute your corner points back into every single constraint inequality to ensure they are fully within the valid region before calculating the final table values.

Answer 1

Concept: To solve a Linear Programming Problem graphically, we map out the boundary lines corresponding to each inequality constraint. By evaluating test points, we can determine the overlapping feasible region and then find the coordinates of its corner points. The optimal solution will always occur at one or more of these corner points.

Step 1: Identify the intersecting boundary lines and determine the feasible region.
Let us convert the given inequalities into boundary lines to find their coordinate intercepts: 1. Line 1: $x + 2y = 40$. Intercepts are at $(40, 0)$ and $(0, 20)$. The test point $(0,0)$ satisfies $0 \le 40$, so the half-plane faces toward the origin. 2. Line 2: $3x + y = 30$. Intercepts are at $(10, 0)$ and $(0, 30)$. The test point $(0,0)$ fails $0 \ge 30$, so the half-plane faces away from the origin. 3. Line 3: $4x + 3y = 60$. Intercepts are at $(15, 0)$ and $(0, 20)$. The test point $(0,0)$ fails $0 \ge 60$, so the half-plane faces away from the origin. By plotting these lines on a coordinate plane alongside the non-negativity constraints $x \ge 0, y \ge 0$, we discover a bounded feasible region.

Step 2: Calculate the exact corner points and evaluate the objective function $Z$.
Let us find the coordinates of the corner points bounding this region by solving the equations of intersecting lines: * Point P1: Intersection of $3x + y = 30$ and $4x + 3y = 60$. Multiply the first equation by 3: $9x + 3y = 90$. Subtract the second equation: $5x = 30 \implies x = 6$. Substitute back: $3(6) + y = 30 \implies y = 12$. So, $P_1 = (6, 12)$. * Point P2: Intersection of $3x + y = 30$ and $x + 2y = 40$. From the first equation, $y = 30 - 3x$. Substitute into the second: $x + 2(30 - 3x) = 40 \implies -5x = -20 \implies x = 4$. Substitute back: $y = 30 - 3(4) = 18$. So, $P_2 = (4, 18)$. * Point P3: Intersection of $x + 2y = 40$ and the $y$-axis ($x=0$), which is $(0, 20)$. Now, let us evaluate the objective function $Z = 20x + 10y$ at each of these corner points: {|c|c|} Corner Point $(x, y)$ & Value of $Z = 20x + 10y$
$(0, 20)$ & $Z = 20(0) + 10(20) = 200$
$(4, 18)$ & $Z = 20(4) + 10(18) = 80 + 180 = 260$
$(6, 12)$ & $Z = 20(6) + 10(12) = 120 + 120 = 240$
*Correction review check:* Let's check if the lines allow any other corner points. The constraints are $x+2y \le 40$, $3x+y \ge 30$, $4x+3y \ge 60$. Let's double-check point $P_1$: $4(6) + 3(12) = 24 + 36 = 60$. Correct. $3(6)+12=30$. Correct. $6+2(12) = 30 \le 40$. Correct. Let's check point $(15,0)$: $15+2(0)=15 \le 40$, $3(15)+0 = 45 \ge 30$, $4(15)+3(0)=60 \ge 60$. Let's test $Z(15,0) = 20(15) = 300$. Let's look closely at the minimum calculation values. Comparing $200$, $260$, and $240$: the absolute lowest value is $200$ at the point $(0, 20)$.