Q.Solve the following linear programming problem graphically: Maximise $Z = 4x + y$ subject to the constraints: $x + y \le 50$, $3x + y \le 90$, $x \ge 0,\ y \ge 0$.

Concept: Linear Programming Graphical Method – The feasible region is the intersection of all constraints; the maximum of $Z$ occurs at a corner point of this region.

Steps:

1. Plot constraints

   $x + y = 50$ passes through $(50,0)$ and $(0,50)$.

   $3x + y = 90$ passes through $(30,0)$ and $(0,90)$.

   Non-negativity restricts to the first quadrant.

2. Find corner points

   Intersection of $x+y=50$ and $3x+y=90$: subtract to get $2x = 40 \Rightarrow x = 20$, then $y = 30$.

   Other corners: $(0,0)$, $(30,0)$, $(0,50)$.

3. Evaluate $Z = 4x + y$

   • $(0,0)$: $Z = 0$
   • $(30,0)$: $Z = 120$
   • $(0,50)$: $Z = 50$
   • $(20,30)$: $Z = 4(20) + 30 = 110$

The maximum value is $120$ at $(30,0)$.

We have a linear programming problem with two decision variables, $x$ and $y$, and a linear objective function $Z = 4x + y$ to be maximised. The constraints are all linear inequalities. The graphical method works here because with only two variables, each inequality describes a half-plane in the $xy$-plane. The intersection of all these half-planes is the feasible region — the set of all points that satisfy every constraint. The fundamental theorem of linear programming tells us that if an optimal solution exists, it will be found at one of the corner points (vertices) of this feasible region. So our job is to draw the region, find its vertices, and evaluate $Z$ at each vertex.

Let’s go step by step.

1. Plot each constraint as a line.

   First, treat each inequality as an equation.

   • $x + y = 50$: This line passes through $(50,0)$ and $(0,50)$.
   • $3x + y = 90$: This passes through $(30,0)$ and $(0,90)$.
   • $x = 0$ is the $y$-axis.
   • $y = 0$ is the $x$-axis.

   The inequalities $x \ge 0$, $y \ge 0$ restrict us to the first quadrant.

2. Determine which side of each line is feasible.

   For $x + y \le 50$, test the origin $(0,0)$: $0 + 0 \le 50$ is true, so the half-plane containing the origin is feasible.

   For $3x + y \le 90$, test $(0,0)$: $0 \le 90$ is true, so again the origin side is feasible.

   So the feasible region is the intersection of the two half-planes below both lines, in the first quadrant.

3. Find the corner points of the feasible region.

   The region is a polygon bounded by the axes and the two lines. The vertices are:

   • $(0,0)$ — intersection of $x=0$ and $y=0$.
   • $(0,50)$ — intersection of $x=0$ and $x+y=50$. But check if it satisfies $3x+y \le 90$: $3(0)+50 = 50 \le 90$, yes.
   • $(30,0)$ — intersection of $y=0$ and $3x+y=90$. Check $x+y \le 50$: $30+0 = 30 \le 50$, yes.
   • The intersection of the two lines $x+y=50$ and $3x+y=90$. Solve: subtract the first from the second: $(3x+y) - (x+y) = 90 - 50$ gives $2x = 40$, so $x = 20$. Then $y = 50 - x = 30$. So the point is $(20,30)$. Check both constraints: $20+30=50$ (tight), $3(20)+30=60+30=90$ (tight). This is inside the first quadrant.

   So the vertices are: $A(0,0)$, $B(0,50)$, $C(20,30)$, $D(30,0)$.

4. Evaluate $Z = 4x + y$ at each vertex.

   Vertex	$x$	$y$	$Z = 4x + y$
   $A$	0	0	$0$
   $B$	0	50	$0 + 50 = 50$
   $C$	20	30	$80 + 30 = 110$
   $D$	30	0	$120 + 0 = 120$

   The largest value is $120$ at $D(30,0)$.

5. Interpret the result. The maximum value of $Z$ is $120$, achieved at $x = 30$, $y = 0$. This means that under the given constraints, the best strategy is to use all resources to produce $x$ (30 units) and none of $y$.

Method: Corner-Point (Graphical) Method for a Maximum

This is the standard technique for maximising a linear objective $Z=ax+by$ of two variables under linear constraints, and it is the only optimisation method in the Class-12 syllabus (no simplex/algebra needed).

Steps

Step 1: Convert each inequality to a line and plot it.

For a constraint $px+qy\le r$, draw $px+qy=r$ using its intercepts $\left(\tfrac{r}{p},0\right)$ and $\left(0,\tfrac{r}{q}\right)$. Do this for every constraint, and keep $x\ge 0,\ y\ge 0$ (first quadrant).

Step 2: Shade the correct half-plane for each constraint.

Substitute a test point (the origin is easiest when the line does not pass through it). If the inequality holds there, keep the origin's side; otherwise keep the other side. The feasible region is the intersection of all the kept half-planes.

Step 3: Find every corner (vertex) of the feasible region.

A vertex is where two boundary lines meet. Solve those two equations simultaneously (e.g. subtract one from another to eliminate a variable), and confirm the point satisfies all the other constraints before accepting it.

Step 4: Evaluate $Z=ax+by$ at each corner and pick the largest.

The vertex giving the biggest value is the optimal solution.

Why the answer need not be at the "inner" intersection: the maximum lands on whichever corner is farthest in the direction the objective grows. If $Z$ increases faster in $x$ than in $y$, a corner on an axis can beat the intersection of two slanted lines — so you must always test every vertex, never guess.

Common Mistakes

Mistake 1: Assuming the maximum is at the intersection of the two slanted lines $(20,30)$.

Why it's wrong: for $Z=4x+y$ the objective grows much faster in $x$, so the true maximum is at $(30,0)$ where $Z=120$, beating $Z=110$ at $(20,30)$. Correct approach: evaluate $Z$ at every vertex and compare — never assume the "inner" corner wins.

Mistake 2: Forgetting the axis-intercept corners $(0,50)$ and $(30,0)$.

Why it's wrong: the feasible polygon includes vertices on the axes, and the optimum here is one of them. Missing a corner can hide the real maximum. Correct approach: list all vertices, including the ones where a constraint meets an axis.

Mistake 3: Shading the wrong side of a $\le$ line.

Why it's wrong: $x+y\le 50$ and $3x+y\le 90$ both keep the origin side; shading outward would give an empty or wrong region. Correct approach: test $(0,0)$ in each inequality — if it holds, keep the origin's side.

Mistake 4: Reporting only $Z=120$ without stating where it occurs.

Why it's wrong: a graphical LP answer must give both the optimal value and the point $(30,0)$. Correct approach: always state "maximum $Z=120$ at $(30,0)$".