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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 for minimisation by graphical method :
Objective function
$
\begin{aligned}Z = 5 x + y \\
constraints
3 x + 5 y & \geq 1 5 \\
5 x + 2 y & \leq 1 0 \\
x \geq 0 , y & \geq 0
\end{aligned}
$
✓

Answer

$
Z=5 x+y
$
Constraints are:
$
\begin{aligned}
3 x+5 y & \geq 15 \\
5 x+2 y & \leq 10 \\
x & \geq 0 \\
y & \geq 0
\end{aligned}
$
(i) Region of $3 x+5 y \geq 15$ :
Image
The line $3 x+5 y=15$ passes through the points $A (5,0)$ and $B (0,3)$. Its graph is AB .
Putting $x=0, y=0$ in $3 x+5 y \geq 15$, we get $0 \geq 15$ which is false.
i.e., this region contain AB and region above it.
(ii) Region of $5 x+2 y \leq 10$ :The line $5 x+2 y=10$ passes through the points P $(2,0)$ and $Q (0,5)$. Its graph is PQ .
Now putting $x=0, y=0$ is $5 x+2 y \leq 10$, we get $0 \leq 10$ which is true.
i.e., the region of $5 x+2 y \leq 10$ is the line PQ and below PQ towards the origin.
(iii) The area of $x \geq 0$ is on $y$-axis and to the right of $y$-axis.
(iv) The area of $y \geq 0$ is on $x$-axis and above the $x$-axis.
Thus, the feasible region of this problem is OBRP.
Corner PointCorresponding Value of Z = 5x + y
O(0, 0)0
P(2, 0)10
$R \left(\frac{20}{19}, \frac{45}{19}\right)$$\frac{ 1 4 5 }{ 1 9 }$ Maximum
B(0 , 3)3

Hence, at the corner point O (0, 0) value of Z = 0

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