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Linear programming — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsLinear programming5 Marks
Question
Solve the following LPP graphically:
Manimize Z = 6x + 3y
Subject to the constraints:
$4\text{x}+\text{y}\geq80$
$\text{x}+5\text{y}\geq115$
$3\text{x}+2\text{y}\leq150$
$\text{x}\geq0,\text{y}\geq0$
✓

Answer

The given contraints are:
$4\text{x}+\text{y}\geq80$
$\text{x}+5\text{y}\geq115$
$3\text{x}+2\text{y}\leq150$
$\text{x}\geq0,\text{y}\geq0$
Converting the given inequation into equation, we get
4x + y = 80, x + 5y = 115, 3x + 2y = 150, x = 0 and y = 0
These lines are drawn on the graph and the shaded region ABC represents the feasible region of the given LPP.

It can be observed that the feasible region is bounded.
The coordinates of the corner points of the feasible region are A(2, 72), B(15, 20), and C(40, 15).
The values of the objective function, Z at these corner points are given in the following table:
Corner point Value of the objective Function Z = 6x + 3y
A(2, 72) Z = 6 × 2 + 3 × 72 = 228
B(15, 20) Z = 6 × 15 + 3 × 20 = 150
C(40, 15) Z = 6 × 40 + 3 × 15 = 150
From the table, Z is manimum at x = 15 and y = 20 and the manimum value of Z is 150.
Thus, the manimum value of Z is 150.

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