Answer the question [ 4 mark question ]

Question 14 Marks

Maximize $Z=3 x+2 y$ subject to constraints $x+2 y \leq 10,3 x+y \leq 15, x \geq 0, y \geq 0$ by using graphical method.

Answer

According to question,
$
\begin{aligned}
Z & =3 x+2 y \\
\text { Constrants, } \quad x+2 y & \leq 10 \\
3 x+y & \leq 15 \\
x, y & \geq 0
\end{aligned}
$

The shaded region in the figure iscoherent region determined by the system of given constraints. We observe that the feasible region OPRBO is bounded. Therefore we will use the corner point method to find the value of Z.

Corner Points	Corresponding value of Z
Ο (0, 0)	Z = 3 * 0 + 2 * 0 = 0
P (5,0)	Z = 3 * 5 + 2 * 0 = 15
R (4, 3)	Z = 3 * 4 + 2 * 3 =18 Maximum
B (0, 5)	Z = 3 * 0 + 2 * 5 = 10

The coordinates of corner points O, P, R and B are (0, 0), (5, 0), (4, 3) and (0, 5) respectively. The maximum value of Z is 18 at point R (4, 3).

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Question 24 Marks

Maximise $z=3 x+5 y$ subject to the constraints -
$\begin{array}{l}3 x+5 y \leq 15 \\5 x+2 y \leq 10 \\x \geq 0, y \geq 0\end{array}$

Answer

Constraints: $3 x+5 y \leq 15,5 x+2 y \leq 10, x, y \geq 0$
Intersection point: Solve $3 x+5 y=15$ and $5 x+2 y=10$
$\Longrightarrow x=20 / 19, y=$ 45/19.
Corner points: $(0,0),(2,0),(0,3),(20 / 19,45 / 19)$.
Evaluate $Z$ :
$\therefore$ At $(0,3), Z=3(0)+5(3)=15$
$\therefore$ At $(20 / 19,45 / 19)$,
$\therefore$ $Z=3(20 / 19)+5(45 / 19)$
$=285 / 19=15$
$\therefore$ Maximum value is 15. It occurs at any point on the line segment joining $(0,3)$ and $(20 / 19$, 45/19).

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Question 34 Marks

Maximise $z=4 x+y$ subject to the constraints -
$\begin{array}{l}x+y \leq 50 \\3 x+y \leq 20 \\x \geq 0, y \geq 0\end{array}$

Answer

Constraints are $x+y \leq 50,3 x+y \leq 20, x, y \geq 0$
Corner points are $O(0,0), A(20 / 3,0)$, and $B(0,20)$.
Evaluate $Z$ at corner points:
$\therefore$ $\operatorname{At}(0,0), Z=0$
$\therefore$ At $(20 / 3,0)$,
$Z=4(20 / 3)+0$
$=20$
$\therefore$ At $(0,20), Z=4(0)+20=20$
Maximum value of Z is 26.67 at point (20/3, 0).

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Question 44 Marks

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 $3 x+5 y \geq 15$ :

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 Point	Corresponding 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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Question 54 Marks

Solve the following linear programming problem by graphical method. Under the following constraints :
$
\begin{aligned}
x+2 y & \geq 10 \\
x+y & \geq 6 \\
3 x+y & \geq 8 \\
x, y & \geq 0
\end{aligned}
$
$\operatorname{minimise} Z=3 x+5 y$.

Answer

On drawing all the inequalities on the graph paper, ABCD is the feasible region of this problem where coordinates are as follows :
$A (0,8), B (1,5), C (2,4)$ and $D (10,0)$
Now we shall find the values of $Z$ at these points according to the following table :

Corner Point	Corresponding Value of Z = 3x + 5y
A(0, 8)	40
B (1, 5)	28
C (2, 4)	26 Minimum
D(10, 0)	30

Hence the minimum value of Z at the corner point C (2, 4) = 26.

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Question 64 Marks

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.

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 Point	Value 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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Question 74 Marks

Maximize $Z=4 x+y$ subject to constraints $x+y$ $\leq 50,3 x+y \leq 90, x \geq 0, y \geq 0$ by using graphical method.

Answer

According to question,
z = yx + y
The constraints are
x+y d50, 3x + y d90, x t0, y to
The shaded ragion in the figure is the coherent region determined by the system of given constraints. We observe that the coherent region OABC is bounded. Therefore we will use the corner point method to find the maximum value of z.

Corner Points	Corresponding value of Z
(0,0)	Z = 4x + y = 4 × 0 + 0 = 0
(30,0)	Z = 4x + y = 4 x 30 + 0 = 120 Maximum
(20, 30)	Z = 4x + y = 4 × 20 + 30 = 110
(0,50)	Z = 4x + y = 4 × 0 + 50 = 50

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