Linear Programming Miscellaneous Exercise Class 12 NCERT Solutions

Understanding Advanced LPP Formulations

Linear Programming (LPP) deals with optimizing (maximizing or minimizing) a linear objective function subject to a set of linear inequalities called constraints. In the linear programming miscellaneous ex class 12 ncert, you encounter diverse word problems that test your ability to translate written conditions into mathematical models.

Every problem consists of three core structural pillars:

1. Decision Variables ($x$ and $y$): These represent the quantities you need to determine (such as the number of units to produce, or kilograms of food to mix).
2. Objective Function ($Z = ax + by$): The linear function that represents profit (to be maximized) or cost (to be minimized).
3. Constraints: Mathematical inequalities of the form $ax + by \leq c$ or $ax + by \geq c$ representing limitations on resources like budget, materials, or nutrition. The non-negativity constraints $x \geq 0, y \geq 0$ are always present. Solving these requires finding the overlapping area of all constraints, called the Feasible Region.

Systematic Process to Solve Miscellaneous LPPs

1. Identify and Define Variables — Read the problem statement carefully. Identify the two primary unknowns that you need to optimize and denote them as $x$ and $y$. State their units clearly (e.g., let $x$ be the number of model A machines and $y$ be the number of model B machines).
2. Construct the Constraint Inequalities — Create a data table grouping parameters if necessary. Set up linear inequalities based on constraints like availability of hours, raw material limits, or nutrient minimums. Always include the non-negativity conditions: $x \ge 0, y \ge 0$.
3. Formulate the Objective Function — Locate the quantity that needs to be optimized (maximum profit, minimum cost, minimum distance). Express it as a function of $x$ and $y$ in the form $Z = ax + by$.
4. Plot the Feasible Region — Convert constraints to linear equations to draw boundary lines. Identify the correct half-planes by testing a point (usually the origin $(0,0)$). Shade the common intersecting area, which is the feasible region.
5. Locate Corner Points and Evaluate Z — Identify all vertex coordinates of the shaded feasible region by solving intersecting line equations. Calculate the value of $Z$ at each corner point. For bounded regions, the highest/lowest $Z$ value is the final answer.

Detailed Solved NCERT Miscellaneous Problems

• Example 1 (Diet Problem): A dietitian wishes to mix two types of food $F_1$ and $F_2$ in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 10 units of vitamin C. Food $F_1$ contains 2 units/kg of vitamin A and 1 unit/kg of vitamin C. Food $F_2$ contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs ₹50/kg to purchase Food $F_1$ and ₹70/kg to purchase Food $F_2$. Formulate this LPP to minimize cost. Step 1: Variables Formulation Let $x$ be the quantity (in kg) of Food $F_1$ and $y$ be the quantity (in kg) of Food $F_2$. Step 2: Linear Constraints - For Vitamin A: $2x + y \geq 8$ - For Vitamin C: $x + 2y \geq 10$ - Non-negativity constraints: $x \geq 0, y \geq 0$ Step 3: Objective Function (Cost Minimized) Minimize $Z = 50x + 70y$ Step 4: Find Corner Points The boundary lines $2x + y = 8$ and $x + 2y = 10$ intersect at $(2, 4)$. The feasible region is unbounded with corner points: $A(0, 8)$, $B(2, 4)$, and $C(10, 0)$. Step 5: Evaluate Cost - At $A(0, 8)$: $Z = 50(0) + 70(8) = ₹560$ - At $B(2, 4)$: $Z = 50(2) + 70(4) = 100 + 280 = ₹380$ - At $C(10, 0)$: $Z = 50(10) + 70(0) = ₹500$ Final Answer: Since the region is unbounded, we verify if $50x + 70y < 380$ shares points with the feasible region. It does not. Hence, the minimum cost is ₹380 when 2 kg of Food $F_1$ and 4 kg of Food $F_2$ are mixed.
• Example 2 (Manufacturing Problem): A manufacturer makes two products, A and B. Product A requires 2 hours of processing on Machine I and 6 hours on Machine II. Product B requires 4 hours on Machine I and 3 hours on Machine II. Machine I is available for at most 16 hours, while Machine II is available for at most 18 hours. Profit on Product A is ₹40 and on Product B is ₹50. Find the maximum profit. Step 1: Variable Definition Let $x$ and $y$ represent the number of units of Product A and B produced, respectively. Step 2: Set Constraints - Machine I: $2x + 4y \leq 16 \implies x + 2y \leq 8$ - Machine II: $6x + 3y \leq 18 \implies 2x + y \leq 6$ - Non-negativity: $x \geq 0, y \geq 0$ Step 3: Define Objective Function Maximize Profit $Z = 40x + 50y$ Step 4: Find Intersections & Corner Points Solving equations $x + 2y = 8$ and $2x + y = 6$ gives point $B(\frac{4}{3}, \frac{10}{3})$. The bounded corner points of the shaded region are $O(0,0)$, $A(3,0)$, $B(\frac{4}{3}, \frac{10}{3})$, and $C(0,4)$. Step 5: Evaluate Profit - At $O(0,0)$: $Z = 0$ - At $A(3,0)$: $Z = 40(3) + 0 = ₹120$ - At $C(0,4)$: $Z = 0 + 50(4) = ₹200$ - At $B(\frac{4}{3}, \frac{10}{3})$: $Z = 40(\frac{4}{3}) + 50(\frac{10}{3}) = \frac{160 + 500}{3} = \frac{660}{3} = ₹220$ Final Answer: Maximum profit is ₹220 achieved by producing $\frac{4}{3}$ units of A and $\frac{10}{3}$ units of B.

Exam Pitfalls & Board Strategy for LPP

To score full marks on the CBSE Board Exam, keep these crucial details in mind:

1. Unbounded Regions Warning: If your feasible region is open (unbounded) and you are finding a minimum value $M$, you MUST draw the dotted line of the open half-plane $ax + by < M$. If this half-plane shares any common point with your feasible region, the minimum value does not exist.
2. Correct Inequality Signs: Look out for keywords: 'At least' means $\geq$, 'At most' or 'Not more than' means $\leq$.
3. Neat Graph Work: Always draw your graph on a clean grid using a pencil and ruler. Clearly label the axes, boundary equations on respective lines, shaded feasible region, and coordinates of all corner points.

Practice Questions with Solutions

• Q: Minimize $Z = 3x + 5y$ subject to constraints: $x + 3y \geq 3$, $x + y \geq 2$, and $x, y \geq 0$. A: Step 1: Write down equations for the constraints: $x + 3y = 3$ and $x + y = 2$. Step 2: Find intersecting points of boundary lines. Solving $x + 3y = 3$ and $x + y = 2$ gives $2y = 1 \implies y = 0.5$ and $x = 1.5$. The intersection point is $P(1.5, 0.5)$. Step 3: Determine the corner points of the unbounded feasible region: $A(0, 2)$, $P(1.5, 0.5)$, and $B(3, 0)$. Step 4: Evaluate the objective function $Z = 3x + 5y$ at each point: - At $A(0, 2)$: $Z = 3(0) + 5(2) = 10$ - At $P(1.5, 0.5)$: $Z = 3(1.5) + 5(0.5) = 4.5 + 2.5 = 7$ - At $B(3, 0)$: $Z = 3(3) + 5(0) = 9$ Step 5: Verify if the open half-plane $3x + 5y < 7$ shares common points with the feasible region. Plotting $3x + 5y < 7$ shows it does not intersect the shaded region. Final answer: The minimum value of $Z$ is 7 at point $(1.5, 0.5)$.
• Q: Maximize $Z = 4x + y$ subject to constraints: $x + y \leq 50$, $3x + y \leq 90$, and $x, y \geq 0$. A: Step 1: Draw the lines $x + y = 50$ and $3x + y = 90$. Step 2: Identify the intersection point of the lines. Subtracting the first equation from the second gives: $2x = 40 \implies x = 20$. Then, $y = 30$. The intersection is $(20, 30)$. Step 3: The bounded feasible region has corner points: $O(0, 0)$, $A(30, 0)$, $B(20, 30)$, and $C(0, 50)$. Step 4: Calculate $Z = 4x + y$ at each corner point: - At $O(0, 0)$: $Z = 0$ - At $A(30, 0)$: $Z = 4(30) + 0 = 120$ - At $B(20, 30)$: $Z = 4(20) + 30 = 80 + 30 = 110$ - At $C(0, 50)$: $Z = 4(0) + 50 = 50$ Final answer: The maximum value of $Z$ is 120, which occurs at the point $(30, 0)$.
• Q: A company produces two items, $X$ and $Y$. The production of item $X$ requires 1 hour of labor and 2 hours of testing. Item $Y$ requires 3 hours of labor and 1 hour of testing. The company has a maximum of 15 labor-hours and 10 testing-hours available per day. If item $X$ yields ₹30 profit and $Y$ yields ₹20, formulate the objective function and constraints. A: Step 1: Define variables. Let $x$ be the number of items of $X$ and $y$ be the number of items of $Y$. Step 2: Determine constraints based on availability limits. - Labor constraints: $1x + 3y \leq 15$ - Testing constraints: $2x + 1y \leq 10$ - Non-negativity constraints: $x \geq 0$, $y \geq 0$ Step 3: Express the objective profit function: Maximize $Z = 30x + 20y$. Final answer: Objective function is Maximize $Z = 30x + 20y$ subject to $x + 3y \leq 15$, $2x + y \leq 10$, and $x, y \geq 0$.
• Q: Maximize $Z = 10x + 6y$ subject to constraints: $x + y \leq 12$, $x \leq 5$, $y \leq 8$, and $x, y \geq 0$. A: Step 1: Plot the constraints $x + y = 12$, $x = 5$, and $y = 8$. Step 2: Find all corner points of the bounded feasible region bounded by these inequalities. The corner points are: $O(0,0)$, $A(5,0)$, $B(5,7)$ (since $x=5$ intersects $x+y=12$ at $y=7$), $C(4,8)$ (since $y=8$ intersects $x+y=12$ at $x=4$), and $D(0,8)$. Step 3: Evaluate $Z = 10x + 6y$ at these corner points: - At $O(0,0)$: $Z = 0$ - At $A(5,0)$: $Z = 10(5) + 0 = 50$ - At $B(5,7)$: $Z = 10(5) + 6(7) = 50 + 42 = 92$ - At $C(4,8)$: $Z = 10(4) + 6(8) = 40 + 48 = 88$ - At $D(0,8)$: $Z = 10(0) + 6(8) = 48$ Final answer: The maximum value of $Z$ is 92 at the corner point $(5, 7)$.

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