Maharashtra State BoardHSC Commerce 12th Board Exam
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Solve the following problem : Maximize Z = 6x + 10y Subject to 3x + 5y ≤ 10, 5x + 3y ≤ 15, x ≥ 0, y≥0 - Mathematics and Statistics

Graph

Solve the following problem :

Maximize Z = 6x + 10y Subject to 3x + 5y ≤ 10, 5x + 3y ≤ 15, x ≥ 0, y≥0

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Solution

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

Inequation equation Double intercept form Points (x1, x2) Points (x1, x2)
3x + 5y ≤ 10 3x + 5y = 10 `x/(10/3) + y/(2)` = 1 A `(10/3, 0)` 
B (0, 2)

3(0) + 5(0) ≤ 10
∴ 0 ≤ 10
∴ Origin-side

5x + 3y ≤ 15 5x + 3y = 15 `x/(3) + y/(5)` = 1 C (3, 0)
D (0, 5)
5(0) + 3(0) ≤ 15
∴ 0 ≤ 15
∴ Origin-side
x ≥ 0 x = 0   R.H.S. of Y-axis
y ≥ 0 y = 0     above X-axis


The shaded portion OCEB is the feasible region.
Whose vertices are O(0, 0), C(3, 0), E and B(0, 2)
E is the point of intersection of lines
3x + 5y = 10 …(i)
5x + 3y = 15 …(ii)
∴ By 5(i) – 3(ii), we get
15x + 25y = 50
15x +   9y = 45
  –        –       –  
         16y = 5

∴ y = `(5)/(16)`

Substituting y = `(5)/(16)` in (i), we get

`3x+ 5 xx (5)/(16)` = 10

∴ x = `(45)/(16)`

∴ E`(45/16, 5/16)`
Here, the objective function is Z = 6x + 10y
Now, we will find minimum value of Z as follows:

Feasible points The value of Z = 6x + 10y
O (0, 0) Z = 6(0) + 10(0) = 0
C (3, 0) Z = 6(3) + 10(0) = 18
E`(5/16, 45/16)` Z = `6(45/16) + 10(5/16)` = 20
B (0, 2) Z = 6(0) + 10(2) = 20

∴ Z has maximum value 20 at all points on the line 3x + 5y = 10 between B (0, 2) and E`(45/16, 5/16)`

∴ There are infinite number of optimum solutions of the given LPP.

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APPEARS IN

Balbharati Mathematics and Statistics 2 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 6 Linear Programming
Miscellaneous Exercise 6 | Q 4.03 | Page 104
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