Q.Solve the following LPP : Maximize $Z=2x+5y$ subject to $2x+3y\le6$, $x+y\ge2$, $x,y\ge0$.

Showing the 12 most recent of 28 on this concept.

1. CBSE 2026Set 65/2/11 markMCQ
   Q.For the feasible region shown below, the non-trivial constraints of the linear programming problem are (A) $x+y \le 5,\ x+3y \le 9$ (B) $x+y \le 5,\ x+3y \ge 9$ (C) $x+y \ge 5,\ x+3y \le 9$ (D) $x+y \ge 5,\ 3x+y \le 9$

   ›Reveal solutionSolution

   The feasible region is bounded by two lines that form its upper boundary. By checking which inequalities produce the shaded area (the region below both lines), the correct constraints are $x+y \le 5$ and $x+3y \le 9$, which is option (A).

   In Linear Programming, the graphical method works because each linear constraint cuts the plane into two half-planes — one where the inequality holds, one where it doesn’t. The feasible region is the intersection of all such half-planes. When you’re given a picture of the region, the trick is to identify which side of each boundary line is shaded.

   The two lines visible in the diagram are:

   • $x + y = 5$ (passing through $(5,0)$ and $(0,5)$)
   • $x + 3y = 9$ (passing through $(9,0)$ and $(0,3)$)

   The feasible region is the pentagon-shaped area that lies below both of these lines (since the origin $(0,0)$ is inside the region, and it satisfies $0 \le 5$ and $0 \le 9$). That means the inequalities must be of the “less than or equal to” type.

   Let’s check each option:

   1. Option (A): $x+y \le 5,\ x+3y \le 9$

      The origin satisfies both. The shaded region is below both lines — matches the diagram.

   2. Option (B): $x+y \le 5,\ x+3y \ge 9$

      The origin fails the second inequality ($0 \ge 9$ is false). So the region would not include the origin — contradicts the diagram.

   3. Option (C): $x+y \ge 5,\ x+3y \le 9$

      The origin fails the first inequality ($0 \ge 5$ is false). Again, the origin would be excluded — not the case.

   4. Option (D): $x+y \ge 5,\ 3x+y \le 9$ …

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2. CBSE 2026Set A1 markMCQ
   Q.The maximum value of $Z = 4x + y$ subject to the constraints $x + y \le 50$, $x \ge 0$, $y \ge 0$ is

   (a) $50$

   (b) $250$

   (c) $0$

   (d) none of these

   ›Reveal solutionSolution

   Evaluate $Z$ at the corner points; the maximum is $200$, not listed.

   The feasible region has corner points $(0,0)$, $(50,0)$, $(0,50)$.

   • $Z(0,0) = 0$
   • $Z(50,0) = 4(50) + 0 = 200$
   • $Z(0,50) = 4(0) + 50 = 50$ …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.What is the maximum value of $Z=3x+4y$ subject to the constraints $x+y\le 4$, $x\ge 0$ and $y\ge 0$?

   (a) 12

   (b) 14

   (c) 16

   (d) 19

   ›Reveal solutionSolution

   Evaluating $Z=3x+4y$ at the corner points of the feasible region gives a maximum of $16$ at $(0,4)$.

   The constraints are $x+y\le4$, $x\ge0$, $y\ge0$. This describes a triangular feasible region with corner points where the boundary lines meet:

   • Intersection of $x=0$ and $y=0$: $(0,0)$
   • Intersection of $x+y=4$ and $y=0$: $(4,0)$
   • Intersection of $x+y=4$ and $x=0$: $(0,4)$ …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.The maximum value of the objective function Z = 3x + 4y under the constraints x + y \le 1, x \ge 0, y \ge 0 will be:

   (a) 4

   (b) 5

   (c) 0

   (d) 6

   ›Reveal solutionSolution

   By the corner-point method, evaluate $Z=3x+4y$ at each vertex of the feasible region and pick the largest.

   Feasible region: $x+y\le 1,\ x\ge 0,\ y\ge 0$ is the triangle with vertices $(0,0)$, $(1,0)$, $(0,1)$.

   Evaluate $Z=3x+4y$ at each corner (Fundamental Theorem of LPP — the optimum of a linear objective over a bounded feasible region occurs at a corner point): …

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5. CBSE 2025Set 65/2/11 markMCQ
   Q.A factory produces two products X and Y. The profit earned by selling X and Y is represented by the objective function $Z = 5x + 7y$, where $x$ and $y$ are the number of units of X and Y respectively sold. Which of the following statement is correct? (A) The objective function maximizes the difference of the profit earned from products X and Y. (B) The objective function measures the total production of products X and Y. (C) The objective function maximizes the combined profit earned from selling X and Y. (D) The objective function ensures the company produces more of product X than product Y.

   ›Reveal solutionSolution

   The objective function $Z = 5x + 7y$ is a linear combination of the number of units sold, where the coefficients (5 and 7) are the per-unit profits. Therefore, $Z$ represents the total profit from selling both products, and the goal is to maximize this combined profit. The correct option is (C).

   ---

   The core idea here is what an objective function means in linear programming. In any optimization problem — whether it's profit, cost, distance, or time — the objective function is a single mathematical expression that quantifies what you want to make as large (or as small) as possible.

   Here, the function is $Z = 5x + 7y$.

   The variables $x$ and $y$ stand for the number of units of product X and product Y that are sold. The numbers 5 and 7 are the profit per unit of X and Y respectively. So:

   • If you sell one unit of X, you add ₹5 to $Z$.
   • If you sell one unit of Y, you add ₹7 to $Z$.

   That means $Z$ is simply the total profit from all units sold:

   $Z = (\text{profit per unit of X}) \times (\text{units of X}) + (\text{profit per unit of Y}) \times (\text{units of Y})$.

   Now, in a typical linear programming problem, you are asked to maximize or minimize this $Z$ subject to some constraints (like limited raw materials, labour, or demand). The question here doesn't give constraints — it only asks what the objective function itself represents.

   Let’s examine each option:

   1. Option (A) says the objective function maximizes the difference of the profits from X and Y.

      That would look like $5x - 7y$ or $7y - 5x$ — a subtraction. But here we have addition, so this is wrong.

   2. Option (B) says it measures the total production (i.e., total number of units).

      Total production would be $x + y$ (just adding the counts, ignoring profit). But here each unit is weighted by its profit, so $Z$ is not the count — it’s the profit.

   3. Option (C) says it maximizes the combined profit from selling X and Y. …

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6. CBSE 2025Set 65/4/11 markMCQ
   Q.The corner points of the feasible region of a Linear Programming Problem are $(0, 2)$, $(3, 0)$, $(6, 0)$, $(6, 8)$ and $(0, 5)$. If $Z = ax + by$; $(a, b > 0)$ be the objective function, and maximum value of $Z$ is obtained at $(0, 2)$ and $(3, 0)$, then the relation between $a$ and $b$ is : (A) $a = b$ (B) $a = 3b$ (C) $b = 6a$ (D) $3a = 2b$

   ›Reveal solutionSolution

   In a linear programming problem, if the maximum occurs at two distinct corner points, the objective function is constant along the edge joining them. Here, the maximum at $(0,2)$ and $(3,0)$ forces $2a = 3b$, so the correct relation is $3a = 2b$, which is option (D).

   The key idea in the graphical method of linear programming is that the optimal value of a linear objective function $Z = ax + by$ (with $a, b > 0$) over a convex feasible region always occurs at a corner point. If it occurs at two different corner points, then every point on the line segment joining them also gives the same optimal value — the objective function is constant along that edge.

   Here, the maximum occurs at both $(0,2)$ and $(3,0)$. That means $Z$ has the same value at these two points. Let's work through the reasoning step by step.

   1. Write the objective function at each given point.

      At $(0,2)$: $Z = a(0) + b(2) = 2b$.

      At $(3,0)$: $Z = a(3) + b(0) = 3a$.

   2. Since both give the same maximum value, we equate them:

   $2b = 3a$

   3. Rearrange to find the relation between $a$ and $b$. From $2b = 3a$, we get $3a = 2b$. This is a direct linear relation.

   Watch out

   A common mistake is to stop at $2b = 3a$ and pick an option like $a = 3b$ or $b = 6a$ by misreading the equation. Always check: $2b = 3a$ means $b = \frac{3}{2}a$, not $b = 3a$ or $a = 3b$. The correct form matching the options is $3a = 2b$.

   4. Verify against the options. …

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7. CBSE 2025Set E1 markMCQ
   Q.The maximum value of $Z = 2x + y$ subject to constraints $x + y \leq 35,\ x \geq 0,\ y \geq 0$ is

   (a) $35$

   (b) $105$

   (c) $70$

   (d) $140$

   ›Reveal solutionSolution

   Evaluate $Z$ at the corner points of the feasible region; the maximum is $70$.

   The feasible region defined by $x+y\leq35,\ x\geq0,\ y\geq0$ is a triangle with corner points $(0,0)$, $(35,0)$ and $(0,35)$.

   Evaluate $Z=2x+y$:

   • At $(0,0)$: $Z=0$ …

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8. CBSE 2025Set E1 markMCQ
   Q.The maximum value of $Z = 3x - y$ subject to constraints $x + y \leq 8,\ x \geq 0,\ y \geq 0$ is

   (a) $-8$

   (b) $24$

   (c) $16$

   (d) $8$

   ›Reveal solutionSolution

   Evaluate $Z = 3x - y$ at the corner points; the largest value is $24$ at $(8,0)$.

   The feasible region for $x + y \le 8,\ x \ge 0,\ y \ge 0$ is a triangle with corner points $(0,0)$, $(8,0)$, $(0,8)$. Evaluate $Z = 3x - y$:

   …

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9. CBSE 2025Set ANNUAL1 markMCQ
   Q.What is the minimum value of $2x+y$ subject to $x+2y\ge6$, $x\ge0$, $y\ge0$?

   (i) 3

   (ii) 5

   (iii) 6

   (iv) 0

   ›Reveal solutionSolution

   Evaluate the objective function at the corner points of the (unbounded) feasible region.

   The feasible region for $x+2y\ge6,\ x\ge0,\ y\ge0$ has corner points where the boundary line $x+2y=6$ meets the axes: $(6,0)$ and $(0,3)$ (the region extends unboundedly away from the origin beyond these).

   …

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10. CBSE 2025Set ANNUAL1 markMCQ
    Q.The maximum value of the objective function $Z=3x+4y$ subject to the constraints $x \ge 0$, $y \ge 0$ and $x+y \le 1$ is

    (a) 7

    (b) 4

    (c) 3

    (d) 10

    ›Reveal solutionSolution

    Evaluate $Z$ at each corner of the triangular feasible region and pick the largest.

    The constraints $x\ge0,\ y\ge0,\ x+y\le1$ define a triangle with vertices

    $(0,0),\quad(1,0),\quad(0,1).$

    Evaluate $Z=3x+4y$ at each corner:

    $Z(0,0)=3(0)+4(0)=0,$

    $Z(1,0)=3(1)+4(0)=3,$

    $Z(0,1)=3(0)+4(1)=4.$

    The greatest value is $4$, attained at $(0,1)$.

    …

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11. CBSE 2024Set 65/3/11 markMCQ
    Q.Of the following, which group of constraints represents the feasible region given below (shown in the figure of the question paper)? (A) $x + 2y \le 76,\ 2x + y \ge 104,\ x, y \ge 0$ (B) $x + 2y \le 76,\ 2x + y \le 104,\ x, y \ge 0$ (C) $x + 2y \ge 76,\ 2x + y \le 104,\ x, y \ge 0$ (D) $x + 2y \ge 76,\ 2x + y \ge 104,\ x, y \ge 0$

    ›Reveal solutionSolution

    The problem asks us to identify the set of linear inequalities that define a given feasible region in a graph. By finding the equations of the boundary lines and testing a point (like the origin) to determine the correct inequality direction, we find the constraints are $x + 2y \le 76$, $2x + y \le 104$, $x \ge 0$, and $y \ge 0$. The correct option is (B).

    In Linear Programming, a "feasible region" is the set of all points $(x, y)$ that satisfy all the given constraints simultaneously. Each linear inequality defines a half-plane, and the feasible region is the intersection of these half-planes. When given a graph of a feasible region, we need to reverse this process: identify the boundary lines, find their equations, and then determine the correct inequality sign ($\le$ or $\ge$) for each line based on which side of the line the feasible region lies.

    Here's how we can determine the constraints from the given figure:

    1. Identify the boundary lines and their intercepts.

       The figure shows a feasible region bounded by two lines in the first quadrant. This immediately tells us that the non-negativity constraints $x \ge 0$ and $y \ge 0$ are part of the group.

       Let's identify the intercepts of the two main lines from the figure:

       • Line 1: This line intersects the x-axis at $(76, 0)$ and the y-axis at $(0, 38)$.
       • Line 2: This line intersects the x-axis at $(52, 0)$ and the y-axis at $(0, 104)$.

    2. Determine the equation for each line.

       We can use the intercept form of a linear equation, $\frac{x}{a} + \frac{y}{b} = 1$, where $a$ is the x-intercept and $b$ is the y-intercept.

       • For Line 1 (intercepts $(76, 0)$ and $(0, 38)$):

    $\frac{x}{76} + \frac{y}{38} = 1$

        To clear the denominators, multiply the entire equation by the least common multiple of $76$ and $38$, which is $76$:
    

    $76 \left( \frac{x}{76} \right) + 76 \left( \frac{y}{38} \right) = 76(1)$

    $x + 2y = 76$

    *   **For Line 2 (intercepts $(52, 0)$ and $(0, 104)$):**
    

    $\frac{x}{52} + \frac{y}{104} = 1$

        To clear the denominators, multiply the entire equation by the least common multiple of $52$ and $104$, which is $104$:
    

    $104 \left( \frac{x}{52} \right) + 104 \left( \frac{y}{104} \right) = 104(1)$

    $2x + y = 104$

    3. Determine the inequality for each line.

       The feasible region is the shaded area. We need to determine if the region satisfies $\le$ or $\ge$ for each line. A common method is to pick a test point that is clearly inside the feasible region (or clearly outside) and substitute its coordinates into the line's equation. The origin $(0,0)$ is usually the easiest test point, provided it does not lie on the line itself. In this case, the origin $(0,0)$ is clearly part of the feasible region.

       • For the line $x + 2y = 76$:

         Test the origin $(0,0)$:

         Substitute $x=0, y=0$ into $x + 2y$:

         $0 + 2(0) = 0$.

         Since the origin $(0,0)$ is within the feasible region, the inequality must hold true for $(0,0)$. Comparing $0$ with $76$, we need $0 \le 76$.

         Therefore, the inequality for this line is $x + 2y \le 76$.

       • For the line $2x + y = 104$: …

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12. CBSE 2024Set D1 markMCQ
    Q.The minimum value of $Z = 3x + 5y$ subject to the constraints where $x + y \leq 2,\ x \geq 0,\ y \geq 0$ is

    (a) $16$

    (b) $15$

    (c) $0$

    (d) none of these

    ›Reveal solutionSolution

    The feasible region includes the origin, where $Z = 0$ is the minimum.

    The constraints $x + y \leq 2,\ x \geq 0,\ y \geq 0$ form a triangle with corner points $(0,0), (2,0), (0,2)$. Evaluate $Z = 3x + 5y$: at $(0,0)$, $Z = 0$; …

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