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Model Paper 2 — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsModel Paper 23 Marks
Question
Solve graphically the following linear programming problem:
Maximise $z = 6x + 3y,$
Subject to the constraints:
$4 x+y \geq 80$
$3 x+2 y \leq 150$
$x+5 y \geq 115$
$x > 0, y \geq 0$
✓

Answer

Subject to the constraints are
$4 x+y \geq 80$
$x+5 y \geq 115$
$3 x+2 y \leq 150$
and the non negative constraint $x, y \geq 0$
Converting the given inequations into equations, 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 $\text{ABC}$ represents the feasible region of the given $\text
{LPP.}$
Image
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 \times 2+3 \times 72=228$
$B(15,20): Z=6 \times 15+3 \times 20=150$
$C (40,15): Z=6 \times 40+3 \times 15=285$
From the table, $Z$ is minimum at $x = 15$ and $y = 20$ and the minimum value of $Z$ is $150.$
Thus, the minimum value of $Z$ is $150$

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