Q.Maximise the function $Z = 11x + 7y$, subject to the constraints: $x \le 3$, $y \le 2$, $x \ge 0$, $y \ge 0$.

Concept: Linear Programming Graphical Method – the feasible region is a rectangle bounded by the axes and the lines $x=3$, $y=2$.

Step 1 – Plot constraints

$x \le 3$ and $y \le 2$ with $x, y \ge 0$ give a rectangle with corners at $(0,0)$, $(3,0)$, $(0,2)$, and $(3,2)$.

Step 2 – Evaluate $Z$ at each corner

• $(0,0)$: $Z = 0$
• $(3,0)$: $Z = 11(3) + 7(0) = 33$
• $(0,2)$: $Z = 11(0) + 7(2) = 14$
• $(3,2)$: $Z = 11(3) + 7(2) = 33 + 14 = 47$

Step 3 – Identify maximum

The largest value is $47$ at $(3,2)$.

Why the graphical method works here

Linear programming problems with two variables can be solved visually. Each inequality cuts the plane into two halves — the side that satisfies it is the feasible side. The region where all constraints overlap is called the feasible region. The key theorem says: if a linear objective function has a maximum (or minimum) over a bounded feasible region, it occurs at a corner point (vertex) of that region. So we don't need to check every point — just the corners.

Here, the constraints are:

• $x \le 3$ — a vertical line at $x=3$, feasible to the left.
• $y \le 2$ — a horizontal line at $y=2$, feasible below.
• $x \ge 0$, $y \ge 0$ — the first quadrant.

That's a rectangle with corners at $(0,0)$, $(3,0)$, $(0,2)$, and $(3,2)$. No sloping lines, no tricky intersections — just a box.

Step-by-step solution

1. Plot the constraints and identify the feasible region. Draw the lines $x=3$ and $y=2$. Shade the side that satisfies each inequality. Since all inequalities are "less than or equal to" with non-negativity, the feasible region is the rectangle with vertices:

Every point inside or on the boundary of this rectangle satisfies all constraints.

2. List the corner points. The four vertices are:

3. Evaluate the objective function at each corner.

4. Compare the values. The largest is $47$ at $(3,2)$.

Method: Corner-Point Method on a Rectangular Region

Use this when every constraint is a simple bound ($x\le h$, $y\le k$, $x\ge0$, $y\ge0$), so the feasible region is a rectangle.

Steps

Step 1: Identify the box.

The two upper bounds $x\le h$, $y\le k$ together with the non-negativity conditions enclose the rectangle with corners $(0,0)$, $(h,0)$, $(0,k)$ and $(h,k)$.

Step 2: Evaluate $Z=ax+by$ at all four corners.

Substitute each vertex into the objective and tabulate.

Step 3: Pick the optimum.

With both coefficients $a,b>0$, the maximum falls at the corner farthest from the origin, $(h,k)$; a minimum would fall at $(0,0)$. Still list all four — the reasoning only holds because the region is a first-quadrant rectangle.

Common Mistakes

Mistake 1: Forgetting that $x\ge0,\,y\ge0$ are constraints.

Why it's wrong: without them the region bounded only by $x\le3,\,y\le2$ is not a closed box and the maximum could fail to exist. Correct approach: include the non-negativity conditions so the feasible region is the rectangle $(0,0),(3,0),(0,2),(3,2)$.

Mistake 2: Assuming the bigger coefficient's axis wins.

Why it's wrong: $x$'s coefficient ($11$) exceeds $y$'s ($7$), tempting a guess of $(3,0)$; but the region is a rectangle, so both bounds can be met at once — $(3,2)$ gives $Z=47$, beating $(3,0)$'s $33$. Correct approach: evaluate all four corners; on a rectangle the max is the far corner $(3,2)$.