English
Tamil Nadu Board of Secondary EducationHSC Commerce Class 11

The maximum value of the objective function Z = 3x + 5y subject to the constraints x ≥ 0, y ≥ 0 and 2x + 5y ≤ 10 is - Business Mathematics and Statistics

Advertisements
Advertisements

Question

The maximum value of the objective function Z = 3x + 5y subject to the constraints x ≥ 0, y ≥ 0 and 2x + 5y ≤ 10 is

Options

  • 6

  • 15

  • 25

  • 31

MCQ

Solution

15

Explanation:

2x + 5y = 10

x 0 5
y 2 0

Corner points Z = 3x + 5y
O(0, 0) 0
A(5, 0) 15
B(0, 2) 12

∴ Maximum value is 15

shaalaa.com
Linear Programming Problem (L.P.P.)
  Is there an error in this question or solution?
Chapter 10: Operations Research - Exercise 10.3 [Page 257]

APPEARS IN

Samacheer Kalvi Business Mathematics and Statistics [English] Class 11 TN Board
Chapter 10 Operations Research
Exercise 10.3 | Q 8 | Page 257

RELATED QUESTIONS

Find the feasible solution of the following inequations:

x - 2y ≤ 2, x + y ≥ 3, - 2x + y ≤ 4, x ≥ 0, y ≥ 0


If John drives a car at a speed of 60 km/hour, he has to spend ₹ 5 per km on petrol. If he drives at a faster speed of 90 km/hour, the cost of petrol increases ₹ 8 per km. He has ₹ 600 to spend on petrol and wishes to travel the maximum distance within an hour. Formulate the above problem as L.P.P.


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


Solve the following LPP:

Maximize z = 5x1 + 6x2 subject to 2x1 + 3x2 ≤ 18, 2x1 + x2 ≤ 12, x1 ≥ 0, x2 ≥ 0.


Solve each of the following inequations graphically using XY-plane:

5y - 12 ≥ 0


Solve the following LPP:

Minimize z = 4x + 2y

Subject to 3x + y ≥ 27, x + y ≥ 21, x + 2y ≥ 30, x ≥ 0, y ≥ 0


The feasible region is the set of point which satisfy.


Minimize z = 2x + 4y is subjected to 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0, y ≥ 0 show that the minimum value of z occurs at more than two points


A company produces two types of pens A and B. Pen A is of superior quality and pen B is of lower quality. Profits on pens A and B are ₹ 5 and ₹ 3 per pen respectively. Raw materials required for each pen A is twice as that of pen B. The supply of raw material is sufficient only for 1000 pens per day. Pen A requires a special clip and only 400 such clips are available per day. For pen B, only 700 clips are available per day. Formulate this problem as a linear programming problem.


Solve the following LP.P.

Maximize z = 13x + 9y,

Subject to 3x + 2y ≤ 12,

x + y ≥ 4,

x ≥ 0,

y ≥ 0.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×