Q.A manufacturer of medicines is preparing a production plan of medicines $M_1$ and $M_2$. There are sufficient raw materials available to make $20000$ bottles of $M_1$ and $40000$ bottles of $M_2$, but there are only $45000$ bottles into which either of the medicines can be put. Further, it takes $3$ hours to prepare enough material to fill $1000$ bottles of $M_1$, it takes $1$ hour to prepare enough material to fill $1000$ bottles of $M_2$, and there are $66$ hours available for this operation. The profit is Rs $8$ per bottle for $M_1$ and Rs $7$ per bottle for $M_2$. How should the manufacturer schedule his/her production in order to maximise profit?

Let $x, y$ be the bottles of $M_1, M_2$. Maximise $Z = 8x + 7y$ subject to

$x \le 20000,\ y \le 40000,\ x + y \le 45000,\ 3x + y \le 66000,\ x,y \ge 0.$ …

Step 1 — Identify. Choose how many bottles of $M_1$ and $M_2$ to fill so as to maximise total profit under the raw-material, bottle and time limits.

Step 2 — Set up variables and the objective. Let $x$ = number of bottles of $M_1$ and $y$ = number of bottles of $M_2$. Profit is Rs $8$ per bottle of $M_1$ and Rs $7$ per bottle of $M_2$, so the objective function to maximise is

Step 3 — Mathematical formulation (the constraints). Raw material caps each medicine: $x \le 20000$, $y \le 40000$. Only $45000$ bottles exist: $x + y \le 45000$. Time: $3$ h fill $1000$ bottles of $M_1$ and $1$ h fills $1000$ of $M_2$, so the hours used are $\tfrac{3x}{1000} + \tfrac{y}{1000} \le 66$, i.e. $3x + y \le 66000$. With non-negativity,

Step 4 — Solve (corner-point method). The feasible region is the convex polygon $OPQRST$ with vertices

where $R$ is the intersection of $x+y=45000$ and $3x+y=66000$ (subtracting: $2x=21000\Rightarrow x=10500,\ y=34500$). Evaluate $Z=8x+7y$ at each corner:

$Z(O)=0,\quad Z(P)=160000,\quad Z(Q)=202000,$ …

Method: Corner-Point Method for LPP Word Problems

This method solves any linear-programming word problem with two decision variables: translate the wording into constraints and an objective function, then compare $Z$ at the corners of the feasible region.

Steps

Step 1: Define the decision variables

Read the question for the two quantities you are choosing (e.g. "how many of each item to produce") and name them $x$ and $y$. Everything else — profit rates, resource limits — becomes a coefficient or bound on $x,y$.

Step 2: Write the objective function

Identify what is being maximised or minimised (profit, cost, time) and express it as a linear combination

where $p,q$ are the given per-unit rates.

Step 3: Translate every resource limit into a constraint

For each limited resource (raw material, hours, bottles, etc.), write how much of it one unit of $x$ and one unit of $y$ consume, and bound the total by what is available:

Watch the units — a rate given "per 1000 units" or "per hour" must be converted consistently before it can be compared to the total hours/materials available.

Step 4: Graph the constraints and find the feasible region …

Common Mistakes

Mistake 1: Mixing up the time-cost units

Writing the time constraint as $3x + y \le 66$ instead of $3x + y \le 66000$. Why it's wrong: "$3$ hours per $1000$ bottles" means $\dfrac{3x}{1000}$ hours are used per $x$ bottles of $M_1$, so the constraint $\dfrac{3x}{1000}+\dfrac{y}{1000}\le 66$ must be cleared of the denominator by multiplying through by $1000$, giving $3x+y\le66000$ — not $66$. Correct approach: always convert "per $1000$ units" rates to the same scale as $x,y$ before writing the inequality.

Mistake 2: Skipping a "less obviously optimal" corner

Evaluating $Z$ only at corners that look far from the origin, skipping intermediate vertices like $Q(20000,6000)$. Why it's wrong: the Corner-Point Theorem guarantees the optimum is at some vertex, but doesn't say which one — any omitted vertex risks missing the true maximum. Correct approach: list every vertex of the feasible polygon systematically (by walking its boundary) and evaluate $Z$ at each one, no exceptions. …