Q.A company makes 3 models of calculators: A, B and C at factory I and factory II. The company has orders for at least 6400 calculators of model A, 4000 calculators of model B and 4800 calculators of model C. At factory I, 50 calculators of model A, 50 of model B and 30 of model C are made every day; at factory II, 40 calculators of model A, 20 of model B and 40 of model C are made every day. It costs Rs 12000 and Rs 15000 each day to operate factory I and II, respectively. Find the number of days each factory should operate to minimise the operating costs and still meet the demand.

Graphical method. Let $x$ = days factory I runs, $y$ = days factory II runs. Minimise $Z=12000x+15000y$.

Demand constraints (daily output $\times$ days $\ge$ order):

with $x,y\ge 0$.

Feasible corner points: $(0,200),\ (32,120),\ (80,60),\ (160,0)$. …

Set up

Let $x$ = number of days factory I operates and $y$ = number of days factory II operates, $x,y\ge 0$. Running costs are Rs $12000$ and Rs $15000$ per day, so minimise

Daily output and orders:

Corner points

The region is unbounded (all "$\ge$"). Its vertices:

• $5x+4y=640 \cap 3x+4y=480$: subtracting gives $2x=160,\ x=80,\ y=60$ → $(80,60)$; check $5x+2y=520\ge 400$ (ok).
• $5x+4y=640 \cap 5x+2y=400$: subtracting gives $2y=240,\ y=120,\ x=32$ → $(32,120)$; check $3x+4y=576\ge 480$ (ok).
• $5x+2y=400$ on the $y$-axis → $(0,200)$; A and C hold (ok).
• $3x+4y=480$ on the $x$-axis → $(160,0)$; A and B hold (ok).

Evaluate the cost

| Vertex | $Z=12000x+15000y$ |

|---|---| …

Method: Graphical (Corner-Point) Method for a Two-Variable LPP

Once a problem is written as "optimise $Z = ax + by$ subject to linear inequalities in $x$ and $y$," this is the standard CBSE technique for finding the optimum.

Steps

Step 1: Draw every constraint as a line.

Replace each inequality by an equation and plot the line from its axis intercepts. Include $x = 0$ and $y = 0$.

Step 2: Shade the correct half-plane.

Test the origin $(0,0)$ in each inequality: if it is satisfied, keep the origin's side; if not, take the other side. Non-negativity confines you to the first quadrant.

Step 3: Identify the feasible region.

It is the single region where all the shaded half-planes overlap.

Step 4: Find the corner points.

Each vertex is the intersection of two boundary lines, found by solving those two equations together. Keep a candidate only if it satisfies every other constraint -- an intersection that breaks a third inequality is not a corner of the region.

Step 5: Evaluate $Z$ at each corner. …

Common Mistakes

Mistake 1: Writing the demand constraints with $\le$ instead of $\ge$.

Why it's wrong: the orders must be met or exceeded, so each is a lower bound; using $\le$ inverts the region and gives a wrong (often infeasible) answer. Correct approach: $50x+40y\ge 6400$, $50x+20y\ge 4000$, $30x+40y\ge 4800$.

Mistake 2: Confusing per-day output with a total.

Why it's wrong: each factory produces its daily figure every day it runs, so the correct term is (output per day) $\times$ (days). Correct approach: build "daily output $\times$ number of days $\ge$ order," then simplify. …