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Q3:
Solve the following Linear Programming Problems graphically:
3. Maximise $Z = 5x + 3y$ subject to $3x + 5y \leq 15$, $5x + 2y \leq 10$, $x \geq 0$, $y \geq 0$.
Solution :
Step 1: Establishing the Objective Function and Constraints
We are tasked with maximizing the objective function $Z = 5x + 3y$ subject to the following non-negative and linear constraints:
- Constraint 1: $3x + 5y \leq 15$
- Constraint 2: $5x + 2y \leq 10$
- Non-negativity: $x \geq 0, y \geq 0$
Step 2: Determining Boundary Lines and Intercepts
To identify the feasible region, we convert the inequalities into equations to find the line intercepts:
| Equation | $x$-intercept $(y=0)$ | $y$-intercept $(x=0)$ |
|---|---|---|
| $3x + 5y = 15$ | $(5, 0)$ | $(0, 3)$ |
| $5x + 2y = 10$ | $(2, 0)$ | $(0, 5)$ |
Step 3: Finding the Intersection Point of Constraints
Solve the system of linear equations to find the intersection of the two boundary lines [By Algebraic Elimination Method]:
Multiply the first equation by 2 and the second by 5:
$6x + 10y = 30$
$25x + 10y = 50$
Subtract the first from the second: $19x = 20 \implies x = \frac{20}{19} \approx 1.05$
Substitute $x$ into $5x + 2y = 10$: $5(\frac{20}{19}) + 2y = 10 \implies \frac{100}{19} + 2y = \frac{190}{19} \implies 2y = \frac{90}{19} \implies y = \frac{45}{19} \approx 2.37$
Step 4: Graphical Representation
Step 5: Corner Point Evaluation
The feasible region is bounded by the vertices $(0,0)$, $(2,0)$, $(\frac{20}{19}, \frac{45}{19})$, and $(0,3)$ [Per the Corner Point Theorem].
| Corner Point $(x, y)$ | $Z = 5x + 3y$ |
|---|---|
| $(0, 0)$ | $5(0) + 3(0) = 0$ |
| $(2, 0)$ | $5(2) + 3(0) = 10$ |
| $(0, 3)$ | $5(0) + 3(3) = 9$ |
| $(\frac{20}{19}, \frac{45}{19})$ | $5(\frac{20}{19}) + 3(\frac{45}{19}) = \frac{100 + 135}{19} = \frac{235}{19} \approx 12.37$ |
Final Solution: The maximum value of $Z$ is $\frac{235}{19}$ occurring at the point $(\frac{20}{19}, \frac{45}{19})$.
More Questions from Class 12 Mathematics Linear Programming EXERCISE 12.1
- Q1: Solve the following Linear Programming Problems graphically: 1. Maximise $Z = 3x + 4y$ subject to the constraints : $x + y \leq 4$, $x \geq 0$, $y \geq 0$.
- Q2: Solve the following Linear Programming Problems graphically: 2. Minimise $Z = – 3x + 4y$ subject to $x + 2y \leq 8$, $3x + 2y \leq 12$, $x \geq 0$, $y \geq 0$.
- Q4: Solve the following Linear Programming Problems graphically: 4. Minimise $Z = 3x + 5y$ such that $x + 3y \geq 3$, $x + y \geq 2$, $x, y \geq 0$.
- Q5: Solve the following Linear Programming Problems graphically: 5. Maximise $Z = 3x + 2y$ subject to $x + 2y \leq 10$, $3x + y \leq 15$, $x, y \geq 0$.
- Q6: Solve the following Linear Programming Problems graphically: 6. Minimise $Z = x + 2y$ subject to $2x + y \geq 3$, $x + 2y \geq 6$, $x, y \geq 0$. Show that the minimum of $Z$ occurs at more than two points.
- Q7: Solve the following Linear Programming Problems graphically: 7. Minimise and Maximise $Z = 5x + 10y$ subject to $x + 2y \leq 120$, $x + y \geq 60$, $x – 2y \geq 0$, $x, y \geq 0$.
- Q8: Solve the following Linear Programming Problems graphically: 8. Minimise and Maximise $Z = x + 2y$ subject to $x + 2y \geq 100$, $2x – y \leq 0$, $2x + y \leq 200$; $x, y \geq 0$.
- Q9: Solve the following Linear Programming Problems graphically: 9. Maximise $Z = – x + 2y$, subject to the constraints: $x \geq 3$, $x + y \geq 5$, $x + 2y \geq 6$, $y \geq 0$.
- Q10: Solve the following Linear Programming Problems graphically: 10. Maximise $Z = x + y$, subject to $x – y \leq –1$, $–x + y \leq 0$, $x, y \geq 0$.
CBSE Solutions for Class 12 Mathematics Linear Programming
Chapters in CBSE - Class 12 Mathematics
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