Maharashtra State BoardHSC Commerce 12th Board Exam
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Solve the following L.P.P. by graphical method : Minimize: Z = 6x + 2y subject to x + 2y ≥ 3, x + 4y ≥ 4, 3x + y ≥ 3, x ≥ 0, y ≥ 0. - Mathematics and Statistics

Graph

Solve the following L.P.P. by graphical method :

Minimize: Z = 6x + 2y subject to x + 2y ≥ 3, x + 4y ≥ 4, 3x + y ≥ 3, x ≥ 0, y ≥ 0.

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Solution

To draw the feasible region, construct table as follows:

Inequality x + 2y ≥ 3 x + 4y ≥ 4 3x + y ≥ 3
Corresponding equation (of line) x + 2y = 3 x + 4y = 4 3x + y = 3
Intersection of line with X-axis (3, 0) (4, 0) (1, 0)
Intersection of line with Y-axis `(0, 3/2)` (0, 1) (0, 3)
Region Non-origin side Non-origin side Non-origin side

Shaded portion XABCDY is the feasible region,
whose vertices are A(4, 0), B, C and D(0, 3).
B is the point of intersection of the lines x + 4y = 4 and x + 2y = 3
Solving the above equations, we get

x = 2, y = `(1)/(2)`    ∴ B ≡ `(2, 1/2)`

C is the point of intersection of the lines x + 2y = 3 and 3x + y = 3.
Solving the above equations, we get

x = `(3)/(5), y = (6)/(5)`  ∴ C ≡ `(3/5, 6/5)`

Here, the objective function is Z = 6x + 2y
Z at A(4, 0) = 6(4) + 2(0) = 24

Z at B`(2, 1/2) = 6(2) + 2(1/2)` = 12 + 1 = 13

Z at C`(3/5, 6/5) = 6(3/5) + 2(6/5) = (18)/(5)) + (12)/(5)` = 6

∴ Z at D(0, 3) = 6(0) + 2(3) = 6

∴ Z has minimum value 6 at C`(3/5, 6/5)` and D(0, 3).

∴ Z is minimum when, x = `(3/5), y = (6/5)`, Z = 6 and x = 0, y = 3, Z = 6.

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