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Model Paper 9 — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSModel Paper 93 Marks
Question
Solve the Linear Programming Problem graphically:
Maximize $Z = 7x + 10y$ Subject to
$x+y \leq 30000$
$y \leq 12000$
$x \geq 6000$
$x \geq y$
$x, y \geq 0$
✓

Answer

We have to maximize $z = 7x + 10y$
First, we will convert the given inequations into equations, we obtain the following equations:
$x + y = 30000 y = 12000 x = 6000 x = y x = 0 $ and $y = 0$
Region represented by $x + y < = 30000$
The line $x + y = 30000$ meets the coordinate axes at $A(30000, 0)$ and $B(0, 30000)$ respectively.
By joining these points we obtain the line $x + y = 30000$ Clearly $(0, 0)$ satisfies the inequation $x + y < = 30000$
So, the region containing the origin represents the solution set of the inequationx$ y < = 30000$
The line $y = 12000$ is the line that passes through $C(0,12000)$ and parallel to $x-$axis.
The line $x = 6000$ is the line that passes through $(6000, 0)$ and parallel to $y-$axis.
Region represented by $x \geq y$ :
The line $x = y$ is the line that passes through the origin. The points to the right of the line $x = y$ satisfy the inequation $x \geq y$Like
by taking the point$ (-12000, 6000).$
Here, $6000 >- 12000$ which implies $y ; x$ Hence, the points to the left of line $x = y$ will not satisfy the given in equation $x \geq y$
Region represented by $x \geq 0$ and $y \geq 0$:
since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the in equations $x x \geq 0$ and $y \geq 0$
The feasible region determined by subject to the constraints are, $x + y x+y \leq 30000, y \leq 12000, x \geq 6000, x \geq y$,and non-negative restrictions, $x \geq 0$ and $y \geq 0$ are as follows:
Image
The corner points of the feasible region are $D(6000, 0), A(3000, 0), F(18000, 12000)$ and $E(12000, 12000).$
The values of objective function at the corner points are as follows:
Corner point $Z = 7x + 10y$
$D(6000, 0)$ $7 x 6000 + 100 = 42000$
$A(3000, 0)$ $7 x 3000 + 10 \times 021000$
$F(18000, 12000)$ $7 x 18000 + 10 \times 12000 - 246000$
$E(12000, 12000)$ $7 x 12000 + 10 \times 12000 - 204000$
We see that the maximum value of the objective function $Z$ is $246000$ which is at $F(18000,12000)$
that means at $x = 18000$ and $y = 12000$
Thus, the optimal value of objective function $z$ is $246000$

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