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Worked Examples · Example 1
Q.

A furniture manufacture makes two products: chairs and tables. Processing of these products is done on two machines A and B. A chair requires 2 hours on machine A and 6 hours on machine B. A table requires 5 hours on machine A and no time on machine B. There are 16 hours per day available on machine A and 30 hours on machine B. Profit gained by the manufacturer from a chair and a table is Rs. 2 and Rs. 10, respectively. Formulate this problem as a linear programming problem to maximize the total profit of the manufacturer.

The data can be tabulated as follows:

MachineChairTableAvailable time
A2 hours5 hours16 hours
B6 hours030 hours
Profit per unitRs. 2Rs. 10
Andaman Nicobar CbseNCERTSubjective· 3mImportance★★★★★
5% · 1/22 Questions
✓ Free question

This is a resource-allocation linear programming problem. We define variables xx (chairs) and yy (tables), write the objective as Z=2x+10yZ = 2x + 10y, and impose constraints from machine hours: 2x+5y≤162x + 5y \le 16 (machine A), 6x≤306x \le 30 (machine B), with x,y≥0x, y \ge 0. The formulation is complete.

The core idea here is that a manufacturer has limited machine hours (resources) and wants to decide how many chairs and tables to produce so that profit is as large as possible. Linear programming gives us a mathematical way to express this: we choose non-negative quantities of each product, subject to the time limits on each machine, to maximize a linear profit function.

Let’s build it step by step.

  1. Define the decision variables.

    Let xx = number of chairs produced per day.

    Let yy = number of tables produced per day.

    These are the quantities we can control. They cannot be negative, so x≥0x \ge 0, y≥0y \ge 0.

  2. Write the objective function.

    Profit from one chair is Rs. 2, from one table is Rs. 10.

    Total profit ZZ (in Rs.) is:

Z=2x+10yZ = 2x + 10y

We want to maximize ZZ.

  1. Formulate the constraints from machine A. Each chair uses 2 hours on machine A, each table uses 5 hours. Total hours used on A cannot exceed 16.

2x+5y≤162x + 5y \le 16

  1. Formulate the constraints from machine B. Each chair uses 6 hours on machine B, each table uses 0 hours. Total hours on B cannot exceed 30.

6x+0y≤30⇒6x≤306x + 0y \le 30 \quad \Rightarrow \quad 6x \le 30

This simplifies to x≤5x \le 5.

  1. Non-negativity constraints. Already noted: x≥0x \ge 0, y≥0y \ge 0.
Watch out

A common mistake is to forget that tables use zero hours on machine B — that’s fine, it just means machine B doesn’t limit table production. Also, do not write 6x≤306x \le 30 as x≤5x \le 5 without keeping the original form; either is acceptable, but the original form is clearer for the LPP.

  1. Assemble the complete LPP. Maximize

Z=2x+10yZ = 2x + 10y

subject to

2x+5y≤162x + 5y \le 16

6x≤306x \le 30

x≥0,  y≥0x \ge 0, \; y \ge 0

Tip

Notice that the profit per table is five times that per chair, but a table uses more of machine A and none of machine B. The optimal solution will likely involve making as many tables as machine A allows, then using remaining A-time and all of B for chairs. This intuition is exactly what the graphical method or simplex will confirm later.

✓Final answer

The linear programming problem is: maximize Z=2x+10yZ = 2x + 10y subject to 2x+5y≤162x + 5y \le 16, 6x≤306x \le 30, x≥0x \ge 0, y≥0y \ge 0.

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