Q.Refer to Exercise 15. Determine the maximum distance that the man can travel.

This is NCERT Exemplar Exercise 15 (the motorcycle problem): riding at $50$ km/h costs Rs $2$ per km on petrol, riding at $80$ km/h costs Rs $3$ per km, with at most Rs $120$ for petrol and at most $1$ hour of time.

Let $x$ = distance (km) ridden at $50$ km/h and $y$ = distance ridden at $80$ km/h. Maximise the total distance $D = x + y$.

Constraints

Corner points of the feasible region: $(0,0),\ (50,0),\ \left(\frac{300}{7},\frac{80}{7}\right),\ (0,40)$. …

The referenced problem (Exercise 15)

A man rides at $50$ km/h costing Rs $2$ per km on petrol, or at $80$ km/h costing Rs $3$ per km. He has at most Rs 120 for petrol and at most 1 hour of time, and wants the maximum distance he can cover.

Setting up

Let $x$ = distance (in km) ridden at $50$ km/h and $y$ = distance ridden at $80$ km/h. We maximise the total distance

Petrol: the cost is $2x + 3y$, so $2x + 3y \le 120$.

Time: time $=$ distance $\div$ speed, so $\dfrac{x}{50} + \dfrac{y}{80} \le 1$. Multiplying by $400$ gives $8x + 5y \le 400$.

Also $x,y \ge 0$.

Corner points

• Origin: $(0,0)$.
• On the $x$-axis the tighter bound is $8x+5y\le 400 \Rightarrow x=50$: point $(50,0)$.
• On the $y$-axis the tighter bound is $2x+3y\le 120 \Rightarrow y=40$: point $(0,40)$.
• Intersection of $2x+3y=120$ and $8x+5y=400$: multiply the first by $4$ to get $8x+12y=480$; subtracting gives $7y=80$, so $y=\dfrac{80}{7}$ and $x=\dfrac{300}{7}$: point $\left(\dfrac{300}{7},\dfrac{80}{7}\right)$. …

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: Mishandling the time constraint.

Why it's wrong: writing $50x + 80y \le 1$ is dimensionally wrong; even the correct $\dfrac{x}{50}+\dfrac{y}{80}\le 1$ must be cleared to a tidy linear form before plotting. Correct approach: multiply through to get $8x + 5y \le 400$.

Mistake 2: Reporting a single-speed corner as the maximum. …