Name: Linear Programming: Manufacturing Problem
Uploaded: 2019-07-19
Duration: 7 min 3 s
Description: In this video we are going to learn linear programming manufacturing problem comprehensively with the help of real life examples. Check it out!

Question

A pineapple firm produces two products-canned pineapple and canned juice. The specific amounts of material, labour and equipment required to produce each product and the availability of each of those resources are shown in the table given below:

	Canned Pineapple	Canned Juice	Available Resources
Labour (man-hour)	$2$	$3$	$12$
Equipment (Machine hours)	$2.3$	$1$	$6.9$
Material (units)	$1.4$	$1$	$4.9$

Assuming one unit each of canned pineapple and canned juice has profit margin of $₹ 1$ and $₹ 2$ respectively, the problem is to determine the units of the products to be produced so as to maximize the profit. Also determine the maximum profit.

Accepted Answer

Let $x_{1}$ and $x_{2}$ be the number of units of canned juice and canned pineapple respectively.

From the given problem constraints are $3 x_{1} + 2 x_{2} \leq 12$ for labour, $x_{1} + 2.3 x_{2} \leq 6.9$ for equipment, $6.9$ for material and $x_{1}, x_{2} \geq 0$ .

Since canned juice and canned pineapple has profit margins $₹ 2 and ₹ 1$ respectively.

So, objective function is :

Maximize $Z = 2 x_{1} + x_{2}$

Hence, complete formulation of LPP is Maximize $Z = 2 x_{1} + x_{2}$

Subject to

$3 x_{1} + 2 x_{2} \leq 12$

$x_{1} + 2.3 x_{2} \leq 6.9$

$x_{1} + 1.4 x_{2} \leq 4.9$

and $x_{1}, x_{2} \geq 0$ .

Changing inequations into equations we get:

$3 x_{1} + 2 x_{2} = 12 (1)$

$x_{1} + 2.3 x_{2} = 6.9 (2)$

$x_{1} + 1.4 x_{2} = 4.9 (3)$

$x_{1} = 0, x_{2} = 0$ .

From, $(1)$

$x_{1} = 0 \Rightarrow 2 x_{2} = 12 \Rightarrow x_{2} = 6 \Rightarrow (0, 6)$

$x_{2} = 0 \Rightarrow 3 x_{1} = 12 \Rightarrow x_{1} = 4 \Rightarrow (4, 0)$

From $(2)$

$x_{1} = 0 \Rightarrow 2.3 x_{2} = 6.9 \Rightarrow x_{2} = 3 \Rightarrow (0, 3)$

$x_{2} = 0 \Rightarrow x_{1} = 6.9 \Rightarrow (6.9, 0)$

From $(3)$

$x_{1} = 0 \Rightarrow 1.4 x_{2} = 4.9 \Rightarrow x_{2} = 3 .5 \Rightarrow (0, 3.5)$

$x_{2} = 0 \Rightarrow x_{1} = 4.9 \Rightarrow (4.9, 0)$

The value of the objective function $Z = 2 x_{1} + x_{2}$ at corner points of feasible region is as below:

Point $(x, y)$ Value of objective function $Z = 2 x_{1} + x_{2}$

$O (0, 0)$ $0$

$A (0, 3)$ $3$

$B (1.8, 2.2)$ $5.8$

$C (3.3, 1.2)$ $7.8$

$D (4, 0)$ $8$

Maximum value of $Z$ is $8$ when $x_{1} = 4, x_{2} = 0$ .

	Calcium (in unit)	Protein (in unit)	Calories (per unit)	Price (in Rs.)
Food $X$ /unit	$2$	$1$	$0.4$	$12$
Food $Y$ /unit	$0.8$	$1.2$	$1.2$	$20$
Minimum daily requirement	$4$	$4$	$2.4$

Important Questions on Linear Programming Problems

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