Solve the linear programming problem: maximize Z = 3x + 4y subject to constraints

Question

Maximize $Z = 3x + 4y$ subject to the constraints:

x + y \leq 4, \quad x + 3y \leq 6, \quad x \geq 0, \quad y \geq 0

(CBSE 2023, 6 marks)

Solution — Step by Step

$x + y = 4$ : passes through $(4, 0)$ and $(0, 4)$ .

$x + 3y = 6$ : passes through $(6, 0)$ and $(0, 2)$ .

Along with $x = 0$ and $y = 0$ , these form the boundary of the feasible region.

The feasible region is bounded by four vertices:

O = $(0, 0)$ — origin
A = $(4, 0)$ — x-intercept of $x + y = 4$
B = intersection of $x + y = 4$ and $x + 3y = 6$ : subtract to get $2y = 2$ , so $y = 1$ , $x = 3$ . Point B = $(3, 1)$ .
C = $(0, 2)$ — y-intercept of $x + 3y = 6$

Vertex	$Z = 3x + 4y$
$O(0, 0)$	$0$
$A(4, 0)$	$12$
$B(3, 1)$	$9 + 4 = 13$
$C(0, 2)$	$0 + 8 = 8$

\boxed{Z_{\max} = 13 \text{ at } (3, 1)}

Why This Works

The corner point theorem guarantees that if a linear function has an optimum over a convex polygonal region, it occurs at a vertex. This is because a linear function can’t have a maximum in the interior of a convex set — it always increases in some direction until it hits a boundary, and then a corner.

So we only need to check the vertices, not every point in the region. For a problem with $n$ constraints, there are at most $\binom{n}{2}$ potential vertices — a manageable number.

Alternative Method — Iso-profit line

Draw the line $Z = 3x + 4y = c$ for increasing values of $c$ . These are parallel lines with slope $-3/4$ . The maximum $Z$ occurs at the vertex where the line $3x + 4y = c$ last touches the feasible region as $c$ increases.

Visually, slide the line $3x + 4y = c$ upward (increasing $c$ ) until it just touches the feasible region — that touching point is the optimal vertex.

In CBSE boards, this question carries 6 marks. The marking scheme awards: 1 mark for plotting constraint lines, 1 mark for shading the feasible region, 1 mark for finding all vertices, 2 marks for the evaluation table, 1 mark for the final answer. You must draw the graph — solving algebraically without a graph loses 2 marks.

Common Mistake

Students sometimes check only the intercepts $(4,0)$ and $(0,2)$ and miss the intersection point $(3,1)$ . The intersection of two constraint lines often gives the optimal vertex — skipping it means you’ll report $Z = 12$ (at $(4,0)$ ) instead of the correct maximum $Z = 13$ (at $(3,1)$ ). Always find all intersection points within the feasible region.

Question

Solution — Step by Step

Why This Works

Alternative Method — Iso-profit line

Common Mistake

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