Question

The supervisor of a manufacturing plant is trying to determine how many of two parts, Part X and Part Y, are to be produced per day. Each part must be processed in three sections of the plant. The time required for the production along with the profit contribution for each part are given in the following table. Time required (Minutes/Unit) Section 1 Section 2 Section 3 Profit/Unit Part X 50 30 18 $2 Part Y 80 45 22 $3 Available time (minutes) 3,600 2,500 1,200 ​ No more than 60 units of Part X and up to 70 units of Part Y can be produced per day.

​a.What are the variables and the objective function?

b. Develop a linear programming model and solve the model to determine the optimal production quantities of Parts X and Y and solve with graphing. No need to include the graph , but please include the corner points for the feasible region

c. What is the maximum profit

Answer 1

ANSWER::

LPP Formulation:

Decision Variables:

X = units of part X

Y = units of part Y

a)

Objective Function:

The objective of the company is to maximize the profit from the production of both products

Total profit = (units of part X) x (profit per unit of X) + (units of Y)x(profit per unit of Y)

Maximize Profit = max. Z = $2X + $3Y

Subject to:

Availability of time in section 1 is 3600 minutes only	Total time required to produce X and Y <= 3600 minutes 50X + 80Y <= 3600
Availability of time in section 2 is 2500 minutes only	30X + 45Y <= 2500
Availability of time in section 3 is 1200 minutes only	18X + 22Y <= 1200
Not more than 60 units of X can be produced	1X + 0Y <= 60
Up to 70 units of Y can be produced	0X + 1Y <= 70
Non-negativity	X, Y >= 0

b)

Graphical Solution:

Plotting of constraint line on graph:

Let, X-axis represents units of X

Y-axis represents units of Y

Constraint	Equality Equation	X-axis coordinate (Y = 0)	Y-axis coordinate (X = 0)	Feasible area
Section 1	50X +80Y = 3600	(72, 0)	(0, 45)	Towards origin
Section 2	30X + 45Y = 2500	(83.33, 0)	(0, 55.55)	Towards origin
Section 3	18X + 12Y = 1200	(66.67, 0)	(0, 100)	Towards origin
Production of X	X = 60	(60, 0)	Parallel to Y- axis	Towards origin
Production of Y	Y = 70	Parallel to X- axis	(0, 70)	Towards origin

Graph :

The coordinates of Feasible area: (0,0) , (60,0) , (60,7.5), and (0,45)

According the extreme points method the objective value is determined as follows::

Extreme Points	Objective function Value (Z = 2X + 3Y)
(0,0)	Z = 2(0) + 3(0) = 0
(60, 0)	Z = 2(60) + 3(0) = 120
(60, 7.5)	Z = 2(60) + 3(7.5) = 142.5
(0, 45)	Z = 2(0) + 3(45) = 135

c

The maximum value of Z is $142.5 at point (60, 7.5). Thus optimal solution is obtained at B (60, 7.5)

Optimal Solution:

Profit = $142.5

X = 60 and Y = 7.5

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