Q.Solve the following Linear Programming Problem Graphically. Minimize $Z = 18x + 10y$ Subject to $4x + y \geq 20$ $2x + 3y \geq 30$ and $x \geq 0$ and $y \geq 0$

Concept: Linear Programming Formulation — Graphical Method for a Minimization Problem.

Step 1 – Convert constraints to equations and find intercepts

For $4x + y = 20$: intercepts are $(5,0)$ and $(0,20)$.

For $2x + 3y = 30$: intercepts are $(15,0)$ and $(0,10)$.

Step 2 – Identify the feasible region

Since both constraints are $\geq$, the feasible region lies above both lines in the first quadrant. The region is unbounded, but a minimum exists at a corner point.

Step 3 – Find corner points

Intersection of $4x + y = 20$ and $2x + 3y = 30$:

Multiply first by 3: $12x + 3y = 60$. Subtract second: $10x = 30 \Rightarrow x = 3$, then $y = 8$. …

We are asked to minimize $Z = 18x + 10y$ subject to two inequality constraints and non-negativity. The graphical method works beautifully here because we have only two decision variables — we can draw the constraints on the $xy$-plane and see where the objective function is smallest.

The key idea: For a minimization problem with “greater than or equal to” constraints, the feasible region is typically unbounded (it goes off to infinity). The minimum, if it exists, will occur at a corner point of the feasible region — just like in maximization. But we must check that the region is not empty and that the objective doesn’t keep decreasing forever.

Let’s go step by step.

1. Convert each inequality to an equation and draw the lines.

First constraint: $4x + y = 20$

Find intercepts:

• If $x = 0$, then $y = 20$ → point $(0, 20)$
• If $y = 0$, then $4x = 20$ → $x = 5$ → point $(5, 0)$

Second constraint: $2x + 3y = 30$

• If $x = 0$, then $3y = 30$ → $y = 10$ → point $(0, 10)$
• If $y = 0$, then $2x = 30$ → $x = 15$ → point $(15, 0)$

Also, $x \geq 0$ and $y \geq 0$ restrict us to the first quadrant.

2. Determine which side of each line is feasible.

For $4x + y \geq 20$: Test the origin $(0,0)$: $4(0) + 0 = 0 \not\geq 20$ → origin is not feasible. So the feasible side is away from the origin — above the line.

For $2x + 3y \geq 30$: Test $(0,0)$: $0 \not\geq 30$ → again, origin is not feasible. So feasible side is above this line as well.

Thus the feasible region is the intersection of the half-planes above both lines, in the first quadrant. This region is unbounded upward and to the right.

3. Find the corner points of the feasible region.

Corner points occur where:

• A constraint line meets an axis, or
• Two constraint lines intersect.

Intersection of the two lines:

Solve

$4x + y = 20$ … (1)

$2x + 3y = 30$ … (2)

From (1): $y = 20 - 4x$

Substitute into (2): $2x + 3(20 - 4x) = 30$

$2x + 60 - 12x = 30$

$-10x = -30$

$x = 3$

Then $y = 20 - 4(3) = 20 - 12 = 8$

So intersection point is $(3, 8)$.

Axis intercepts that are feasible:

• From $4x + y = 20$: $(5, 0)$ is on the $x$-axis. Check if it satisfies $2x + 3y \geq 30$? $2(5) + 0 = 10 \not\geq 30$ → not feasible. So $(5,0)$ is not a corner of the feasible region.
• From $2x + 3y = 30$: $(15, 0)$ is on the $x$-axis. Check $4x + y \geq 20$: $4(15) + 0 = 60 \geq 20$ → feasible. So $(15, 0)$ is a corner.
• From $4x + y = 20$: $(0, 20)$ is on the $y$-axis. Check $2x + 3y \geq 30$: $0 + 60 = 60 \geq 30$ → feasible. So $(0, 20)$ is a corner.
• From $2x + 3y = 30$: $(0, 10)$ is on the $y$-axis. Check $4x + y \geq 20$: $0 + 10 = 10 \not\geq 20$ → not feasible.

So the feasible corner points are: $(3, 8)$, $(15, 0)$, and $(0, 20)$.

4. Evaluate $Z = 18x + 10y$ at each corner point. …