Q.A manufacturer produces two Models of bikes - Model X and Model Y. Model X takes 6 man-hours to make per unit, while Model Y takes 10 man-hours per unit. There is a total of 450 man-hours available per week. Handling and Marketing costs are Rs 2000 and Rs 1000 per unit for Models X and Y respectively. The total funds available for these purposes are Rs 80,000 per week. Profits per unit for Models X and Y are Rs 1000 and Rs 500, respectively. How many bikes of each model should the manufacturer produce so as to yield a maximum profit? Find the maximum profit.

Concept: Optimization Word Problem (Linear Programming)

Step 1 – Define variables and constraints

Let $x$ = number of Model X bikes, $y$ = number of Model Y bikes.

Man-hours: $6x + 10y \leq 450$

Cost: $2000x + 1000y \leq 80000$

Non-negativity: $x \geq 0,\ y \geq 0$

Step 2 – Objective function

Maximize profit $P = 1000x + 500y$

Step 3 – Find corner points

Simplify constraints:

$3x + 5y \leq 225$ (dividing man-hours by 2)

$2x + y \leq 80$ (dividing cost by 1000)

Solve intersections:

• With $x=0$: $y=45$ from first, $y=80$ from second → feasible point $(0,45)$
• With $y=0$: $x=75$ from first, $x=40$ from second → feasible point $(40,0)$
• Intersection: solve $3x+5y=225$ and $2x+y=80$ → multiply second by 5: $10x+5y=400$; subtract first: $7x=175 \Rightarrow x=25$, then $y=80-2(25)=30$ → point $(25,30)$

Step 4 – Evaluate profit …

Setup. Let $x=$ number of Model X and $y=$ number of Model Y produced per week.

• Man-hours: $6x+10y\le 450\ \Rightarrow\ 3x+5y\le 225$
• Handling/marketing cost: $2000x+1000y\le 80000\ \Rightarrow\ 2x+y\le 80$
• $x\ge 0,\ y\ge 0$
• Objective: maximise $P=1000x+500y$.

Corner points of the feasible region and profit:

• $(0,0)$: $P=0$
• $(40,0)$ - where $2x+y=80$ meets the $x$-axis (man-hours $240\le 450$, ok): $P=40000$
• $(0,45)$ - where $3x+5y=225$ meets the $y$-axis (cost $45\le 80$, ok): $P=22500$
• $(25,30)$ - intersection of $2x+y=80$ and $3x+5y=225$: $P=25000+15000=40000$ …

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: Reporting only one optimal point when there are infinitely many.

Why it's wrong: here the profit line has the same slope as the boundary $2x+y=80$, so the maximum Rs 40,000 is attained all along the edge from $(40,0)$ to $(25,30)$, not at a single corner. Correct approach: when two adjacent corners give the same optimal value, state both and note that every point on the joining segment is also optimal.

Mistake 2: Not recognising the tell-tale sign of multiple optima. …