मराठी
तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी वाणिज्य इयत्ता ११

A company manufactures two models of voltage stabilizers viz., ordinary and auto-cut. All components of the stabilizers are purchased from outside sources, assembly and testing is carried out at the - Business Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

A company manufactures two models of voltage stabilizers viz., ordinary and auto-cut. All components of the stabilizers are purchased from outside sources, assembly and testing is carried out at the company’s own works. The assembly and testing time required for the two models are 0.8 hours each for ordinary and 1.20 hours each for auto-cut. Manufacturing capacity 720 hours at present is available per week. The market for the two models has been surveyed which suggests a maximum weekly sale of 600 units of ordinary and 400 units of auto-cut. Profit per unit for ordinary and auto-cut models has been estimated at ₹ 100 and ₹ 150 respectively. Formulate the linear programming problem.

बेरीज
Advertisements

उत्तर

(i) Variables: Let x1 and x2 denote the number of ordinary and auto-cut voltage stabilized.

(ii) Objective function:

Profit on x1 units of ordinary stabilizers = 100x1

Profit on x2 units of auto-cut stabilized = 150x2

Total profit = 100x1 + 150x2

Let Z = 100x1 + 150x2, which is the objective function.

Since the profit is to be maximized. We have to Maximize, Z = 100x1 + 15x2

(iii) Constraints: The assembling and testing time required for x1 units of ordinary stabilizers = 0.8x1 and for x2 units of auto-cut stabilizers = 1.2x2

Since the manufacturing capacity is 720 hours per week.

We get 0.8x1 + 1.2x2 ≤ 720

Maximum weekly sale of ordinary stabilizer is 600 i.e., x1 ≤ 600

Maximum weekly sales of auto-cut stabilizer is 400 i.e., x2 ≤ 400

(iv) Non-negative restrictions: Since the number of both the types of stabilizers is non-negative, we get x1, x2 ≥ 0.

Thus, the mathematical formulation of the LPP is, Maximize Z = 100x2 + 150x2

Subject to the constraints

0.8x1 + 1.2x2 ≤ 720, x1 ≤ 600, x2 ≤ 400, x1, x2 ≥ 0

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 10: Operations Research - Exercise 10.1 [पृष्ठ २४४]

APPEARS IN

सामाचीर कलवी Business Mathematics and Statistics [English] Class 11 TN Board
पाठ 10 Operations Research
Exercise 10.1 | Q 3 | पृष्ठ २४४

संबंधित प्रश्‍न

A company produces two types of articles A and B which requires silver and gold. Each unit of A requires 3 gm of silver and 1 gm of gold, while each unit of B requires 2 gm of silver and 2 gm of gold. The company has 6 gm of silver and 4 gm of gold. Construct the inequations and find feasible solution graphically.


Solve the following LPP by graphical method:

Maximize z = 4x + 6y, subject to 3x + 2y ≤ 12, x + y ≥ 4, x, y ≥ 0.


The corner points of the feasible solution are (0, 0), (2, 0), `(12/7, 3/7)`, (0, 1). Then z = 7x + y is maximum at ______.


The half-plane represented by 4x + 3y >14 contains the point ______.


Objective function of LPP is ______.


If the corner points of the feasible region are (0, 0), (3, 0), (2, 1) and `(0, 7/3)` the maximum value of z = 4x + 5y is ______.


The constraint that in a particular XII class, number of boys (y) are less than number of girls (x) is given by ______


A firm manufactures two products A and B on which the profits earned per unit are ₹ 3 and ₹ 4 respectively. Each product is processed on two machines M1 and M2. Product A requires one minute of processing time on M1 and two minutes on M2, While B requires one minute on M1 and one minute on M2. Machine M1 is available for not more than 7 hrs 30 minutes while M2 is available for 10 hrs during any working day. Formulate this problem as a linear programming problem to maximize the profit.


The minimum value of z = 5x + 13y subject to constraints 2x + 3y ≤ 18, x + y ≥ 10, x ≥ 0, y ≥ 2 is ______ 


The set of feasible solutions of LPP is a ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×