MHT CET 2026 15th April Evening Shift | Linear Programming Question 12 | Mathematics

The difference between the maximum and minimum values of the objective function $Z = 3x + 5y$, subject to the constraints $x + 3y \leq 60$, $x + y \geq 10$, $x - y \leq 0$, $x, y \geq 0$ is

$50$

$60$

$70$

$80$

MHT CET 2026 15th April Morning Shift

MCQ (Single Correct Answer)

-0

The minimum value of $Z = 3x + y$, subject to the constraints $2x + 3y \leq 6, x + y \geq 1, x \geq 0, y \geq 0$ is....

$5$

$2$

$1$

$9$

MHT CET 2025 5th May Evening Shift

MCQ (Single Correct Answer)

-0

The feasible region for the constraints $x-y \geq 0, x-5 y \leq-5, x \geq 0, y \geq 0$ is shown by the figure:

MHT CET 2025 5th May Evening Shift Mathematics - Linear Programming Question 14 English Option 1

MHT CET 2025 5th May Evening Shift Mathematics - Linear Programming Question 14 English Option 2

MHT CET 2025 5th May Evening Shift Mathematics - Linear Programming Question 14 English Option 3

MHT CET 2025 5th May Evening Shift Mathematics - Linear Programming Question 14 English Option 4

MHT CET 2025 26th April Evening Shift

MCQ (Single Correct Answer)

-0

A manufacturing company produces two items, A and B. Each toy should be processed by two machines, I and II. Machine I can be operated for maximum 10 hours 40 minutes. It takes 20 minutes for an item of A and 15 minutes for B. Machine II can be operated for a total time at 8 hours 20 minutes. It takes 5 minutes for an item A and 8 minutes for B . The profit per item of $A$ is $Rs 25$ and per item of $B$ is ₹ 18 . The formulation of an L.P.P. to maximize the profit (where $x$ is number of items A and $y$ is the number of item $B$ ) is

$$ \begin{aligned} & \text { Maximize } \mathrm{z}=25 x+18 y \\ & \text { subject to } 20 x+15 y \leq 640 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, 5 x+8 y \geq 500 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x, y \geq 0 \end{aligned} $$

Maximize $z=25 x+18 y$

$$ \begin{aligned} \text { subject to } 20 x+15 y & \leq 640 \\ 5 x+8 y & \leq 500 \\ x, y & \geq 0 \end{aligned} $$

$$ \begin{array}{r} \text { Maximize } z=25 x+18 y \\ \text { subject to } 20 x+5 y \leq 8 \\ 5 x+8 y \leq 10 \\ x, y \geq 0 \end{array} $$

$$ \begin{aligned} & \text { Maximize } \mathrm{z}=25 x+18 y \\ & \text { subject to } 4 x+3 y \leq 128 \\ & \qquad \begin{array}{r} 5 x+8 y \geq 500 \\ x, y \geq 0 \end{array} \end{aligned} $$

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