Algebra

Linear Programming

Medium

applied

A furniture trader deals in only two items - chairs and tables. He has Rs. 50,000 to invest and a space to store almost 35 items. A chair costs him Rs. 1000 and a table costs him Rs. 2000. The trader earns a profit of Rs. 150 and Rs. 250 on a chair and a table, respectively. Choose the correct option among following that describes the given linear programming problem (LPP) to maximize the profit ( where x and y are the number of chairs and tables that trader buys and sells)?

Maximize $z = 150x + 250y$ ;

Subjected to constants,

$x + y \leq 35, x + 2y \geq 50, x \geq 0, y \geq 0$

Maximize $z = 150x + 250y$ ;

Subjected to constants,

$x + y \leq 35, x + 2y \leq 50, x \geq 0, y \geq 0$

Maximize $z = 150x + 250y$ ;

Subjected to constants,

$x + y \geq 35, 2x + y \leq 50, x \geq 0, y \geq 0$

Maximize $z = 150x + 250y$ ;

Subjected to constants,

$x + y \geq 35, 2x + y \geq 50, x \geq 0, y \geq 0$

✅ Correct Option: 2

Let $x$ = number of chairs and $y$ = number of tables.

The trader earns a profit of Rs. 150 per chair and Rs. 250 per table.

Total Profit: $z = 150x + 250y$

The objective is to maximize this profit.

The trader can store at most 35 items:

$x + y \leq 35$

The trader has Rs. 50,000 to invest. A chair costs Rs. 1000 and a table costs Rs. 2000:

$1000x + 2000y \leq 50000$

Dividing by 1000:

$x + 2y \leq 50$

The number of chairs and tables cannot be negative:

$x \geq 0$

$y \geq 0$

The complete Linear Programming Problem formulation:

Maximize $z = 150x + 250y$

Subject to:

$x + y \leq 35$

$x + 2y \leq 50$

$x \geq 0, y \geq 0$

Checking the options:

Option 1: Contains $x \leq 0$ , which is incorrect as the number of chairs cannot be negative.

Option 2: Contains all correct constraints: $x + y \leq 35$ , $x + 2y \leq 50$ , $x \geq 0$ , $y \geq 0$

Option 3: Contains $x + y \geq 35$ (minimum requirement) instead of $x + y \leq 35$ (maximum capacity), and has $x \leq 0$ , which is incorrect.

Option 4: Contains $x + y \geq 35$ and $2x + y \geq 50$ , which represent minimum requirements instead of maximum limits.

Therefore, the correct option is Option 2.

2022: 25 May Shift 1

The feasible region of an LPP Max $Z = 3x + 2y$ subject to $x \geq 0, y \geq 0, x - 2y \leq 3$ is:

2025: 21 May Shift 2

About linear programming problem (LPP), which of the following statements are correct?

(A) In a LPP, the linear inequalities or restrictions on the variables are called linear constraints.

(B) If the feasible region for an LPP is unbounded, then the maximum or minimum value of the objective function $Z = ax + by$ never exists.

(D) The common region determined by all the linear constraints of an LPP is called the feasible region.

Choose the correct answer from the options given below:

2025: 29 May Shift 2

The feasible region of a linear programming problem is bounded. The corresponding objective function is Z= 3x-4y. The objective function attains

2025: 2 June Shift 1

The maximum value of $z$ for the linear programing problem maximize $z = x + y$ subject to the constraints $x + 4y \leq 8, 2x + 3y \leq 12, 3x + y \leq 9, x \geq 0, y \geq 0$ is:

2025: 3 June Shift 2

The region represented by the system of inequalities $x, y \geq 0, y \leq 6, x + y \leq 3$

2025: 21 May Shift 2

The corner points of the bounded feasible region determined by the system of linear constraints are $(0, 0)$ , $(5, 0)$ , $(6, 5)$ , $(6, 8)$ , $(4, 10)$ , $(0, 8)$ . Let $Z = 3x - 4y$ be the objective function. The minimum value of Z occurs at

Solution

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