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Linear Programming — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSLinear Programming4 Marks
Question
Solve the following linear programming problem by graphical method.Under the constraints, maximise $Z=60 x+40 y$.
$
\begin{aligned}
x+2 y & \leq 12 \\
2 x+y & \leq 12 \\
x+\frac{5}{4} y & \geq 5 ; x \geq 0, y \geq 0
\end{aligned}
$
✓

Answer

The given constraints are as follows :
$
\begin{aligned}
x+2 y & \leq 12\quad \quad \ldots \ldots(1) \\
2 x+y & \leq 12\quad \quad \ldots \ldots(2) \\
x+\frac{5}{4} y & \geq 5 \quad \quad \ldots \ldots(3)\\
x \geq 0, y & \geq 0\quad \quad \ldots \ldots(4)
\end{aligned}
$
We draw the graph of the constraints (1) to (4). As shown with figure, the feasible region is ABCDE (shaded) which has been determined by the constraints (1) to (4). On observation we find that the feasible region is bounded.
Image
The coordinates of the corner points $A , B , C , D$ and E and respectively $(5,0),(6,0),(4,4),(0,6)$ and $(0,4)$.
Corner PointValue of Z = 60x + 40y
(5,0)300
(6,0)360
(4, 4)400 ← Maximum
(0,6)240
(0,4)1600

We see that at the point (4, 4), the value of Z is maximum.

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