In the following figure, the shaded region represents the system of constraints:

  • A
    $2x + y \leq 12, x + 2y \leq 12, x + 1.25y \geq 5, x \leq 0, y \geq 0$
  • B
    $2x + y \leq 12, x + 2y \leq 12, x + 1.25y \geq 5, x \geq 0, y \leq 0$
  • C
    $2x + y \leq 12, x + 2y \leq 12, x + 1.25y \leq 5, x \geq 0, y \geq 0$
  • D
    $2x + y \leq 12, x + 2y \leq 12, x + 1.25y \geq 5, x \geq 0, y \geq 0$

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

$A$ factory manufactures two types of screws, $A$ and $B$. Each type of screw requires the use of two machines, an automatic and a hand-operated one. It takes $4 \, \text{minutes}$ on the automatic and $6 \, \text{minutes}$ on the hand-operated machine to manufacture a package of screws $A$, while it takes $6 \, \text{minutes}$ on the automatic and $3 \, \text{minutes}$ on the hand-operated machine to manufacture a package of screws $B$. Each machine is available for at most $4 \, \text{hours}$ on any day. The manufacturer can sell a package of screws $A$ at a profit of $Rs. \, 7$ and screws $B$ at a profit of $Rs. \, 10$. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.

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$A$ manufacturing company produces two items,$A$ and $B$. Each item must be processed by two machines,$I$ and $II$. Machine $I$ can be operated for a maximum of $10$ hours $40$ minutes ($640$ minutes). It takes $20$ minutes for an item $A$ and $15$ minutes for an item $B$. Machine $II$ can be operated for a maximum of $8$ hours $20$ minutes ($500$ minutes). It takes $5$ minutes for an item $A$ and $8$ minutes for an item $B$. The profit per item of $A$ is ₹ $25$ and per item of $B$ is ₹ $18$. The formulation of an $L.P.P.$ to maximize the profit (where $x$ is the number of items $A$ and $y$ is the number of items $B$) is . . . . . . .

The shaded area in the figure given below is a solution set of a system of inequations. The minimum value of the objective function $Z = 3x + 5y$,subject to the linear constraints given by this system of inequations,is:

The maximum value of $Z=10 x+25 y$ subject to $0 \leq x \leq 3, 0 \leq y \leq 3, x+y \leq 5, x \geq 0, y \geq 0$ is

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