Maharashtra State BoardHSC Commerce 12th Board Exam
Advertisement Remove all ads

Solve the following problem : Maximize Z = 4x1 + 3x2 Subject to 3x1 + x2 ≤ 15, 3x1 + 4x2 ≤ 24, x1 ≥ 0, x2 ≥ 0 - Mathematics and Statistics

Advertisement Remove all ads
Advertisement Remove all ads
Graph
Sum

Solve the following problem :

Maximize Z = 4x1 + 3x2 Subject to 3x1 + x2 ≤ 15, 3x1 + 4x2 ≤ 24, x1 ≥ 0, x2 ≥ 0

Advertisement Remove all ads

Solution

To find the graphical solution, construct the table as follows:

Inequation Equation Double intercept form Points (x1, x2) Region
3x1 + x2 ≤ 15 3x1 + x2 = 15 `x_1/(5) + x_2/(15)` = 1 A (5, 0)
B (0, 15)
3(0) + 0 ≤ 15
∴ 0 ≤ 15
∴ origin side
3x1 + 4x2 ≤  24 3x1 + 4x2 = 24 `x_1/8 + x_2/(6)` = 1 C (8, 0)
D (0, 6)
3(0 + 4(0) ≤ 24
∴ 0 ≤ 24
∴ origin side
x1 ≥ 0 x1 = 0 R.H.S. of Y-axis
x2 ≥ 0 x2 = 0 above X-axis


Shaded portion ODEA is the feasible region.
Whose vertices are O (0, 0), D (0, 6), E, A (5, 0)
E is the point of intersection of the lines
3x1 + x2 = 15               …(i)
and 3x1 + 4x2 = 24     …(ii)
∴ By (i) – (ii), we get
   3x1 +   x2 = 15
   3x1 + 4x2 = 24
    –    –        –      
          –3x2 = – 9

∴ x2 = `(-9)/(-3)`

∴ x2 = 3
Substituting x2 = 3 in i, we get
3x1 + 3 = 15
∴ 3x1 = 15 – 3
∴ 3x1 = 12
∴ x1 = `(12)/(3)` = 4
∴ E (4, 3)
Here, the objective function is Z = 4x1 + 3x2
Now, we will find maximum value of Z as follows:

Feasible points The value of Z = 4x1 + 3x2
O (0, 0) Z = 4(0) + 3(0) = 0
D (0, 6) Z = 4(0) + 3(0) = 0
E (4, 3) Z = 4(4) + 3(3) = 16 + 9 = 25
A (5, 0) Z = 4(4) + 3(3) = 16 + 9 = 25

∴ Z has maximum value 25 at E(4, 3)
∴ Z is maximum, when x1 = 4, x2 = 3.

Concept: Mathematical Formulation of Linear Programming Problem
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

Balbharati Mathematics and Statistics 2 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 6 Linear Programming
Miscellaneous Exercise 6 | Q 4.05 | Page 104
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×