Maharashtra State BoardHSC Commerce 12th Board Exam
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Solve the following L.P.P. by graphical method : Maximize : Z = 7x + 11y subject to 3x + 5y ≤ 26, 5x + 3y ≤ 30, x ≥ 0, y ≥ 0. - Mathematics and Statistics

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Graph

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

Maximize : Z = 7x + 11y subject to 3x + 5y ≤ 26, 5x + 3y ≤ 30, x ≥ 0, y ≥ 0.

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Solution

The draw the feasible region, construct table as follows:

Inequality 3x + 5y ≤ 26 5x + 3y ≤ 26
Corresponding equation (of line) 3x + 5y  26 5x + 3y = 30
Intersection of line with X-axis `(26/3, 0)` (6, 0)
Intersection of line withY-axis `(0,  26/5)` (0, 10)
Region Origin side Origin side

Shaded portion OABC is the feasible region,

whose vertices are O(0, 0), A(6, 0), B and C `(0,  26/5)`.

B is the point of intersection of the lines 5x + 3y = 30 and 3x + 5y = 26.
Solving the above equations, we get

x = `(9)/(2), y = (5)/(2)`

∴ B = `(9/2 , 5/2)` ≡ (4.5, 2.5)

Here, the objective function is Z = 7x + 11y
∴ Z at O(0, 0) = 7(0) + 11(0) = 0
Z at A(6, 0) = 7(6) + 11(0) = 42

Z at B `(9/2, 5/2) = 7(9/2) + 11(5/2) = (63 + 55)/(2)` = 59

Z at C`(0, 26/5) = 7(0) + 11(26/5)  (286)/(5)` = 57.2

∴ Z has maximum value 59 at B`(9/2, 5/2)`.
i.e. at B(4.5, 2.5)
∴ Z is maximum, when `x = (9)/(2)  "and"  y = (5)/(2)`
i.e. when x = 4.5 and y = 2.5

Concept: Mathematical Formulation of Linear Programming Problem
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