2 Marks Questions

Question 12 Marks

Let R be the feasible region (convex polygon) for a linear programming problem and let Z = ax + by be the objective function. When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities, this optimal value must occur at a corner points (vertex) of the feasible region.
Q. 1. Solve the following LPP graphically:
Max. Z = 2x + 3y
subject to $\begin{aligned} x+y & \leq 4 \\ x & \geq 0, y \geq 0\end{aligned}$
Q. 2. Draw the graph of given LPP and find the corner points of feasible region.
Minimize Z = 200x + 500y
Subject to constraints:
$\begin{aligned} x+2 y & \geq 10 \\ 3 x+4 y & \leq 24 \\ x & \geq 0, y \geq 0\end{aligned}$

Answer

1. Graph of the given LPP is shown as follows.

Corner points of feasible region are:
O(0, 0) A(4, 0) and B(0, 4)

Corner points	Value of Z
0(0, 0)	0
A(4, 0)	8
B(0, 4)	$12 \rightarrow$ Maximum

$\therefore$ Maximum value of Z is 12.
Ans. 2.

The corner points of feasible region are:
A(0, 5) B(0, 6) and B(4, 3).
[Coordinates of point B is obtained by solving equationsx x + 2y = 10 and 3x + 4y = 24 24simultaneously]

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Question 22 Marks

Seema rides her bike at 25 km/h. She has to spend ₹ 2 per km on diesel and if she rides it at faster speed of 40 km/h, the diesel cost increases to ₹ 5 per km. She has 100 to spend on diesel. Let she travels x km with speed 25 km/h and y km with speed 40 km/h. It is given that objective function is Max. Z = x + y.

Q. 1. Construct the LPP and draw the graph for feasible region.
Q. 2. Write the corner points of feasible region and find maximum value of objective function.

Answer

Ans. 1. Required LPP is:
Max Z = x + y
Subject to constraints:
$\begin{aligned} 2 x+5 y & \leq 100 \\ \frac{x}{25}+\frac{y}{40} & \leq 1 \\ x, y & \geq 0\end{aligned}$

Ans. 2. Intersection point of 2x + 5y = 100 and $\frac{x}{25}+\frac{y}{40}=1$ is E.
On solving the equations, we get the coordinates of E are $\left(\frac{50}{3}, \frac{40}{3}\right)$.
So, corner points of LPP are:
O(0, 0) B(0, 20) C(25,0) $E \left(\frac{50}{3}, \frac{40}{3}\right)$

Corner points	Value of Z = x + y
(0,0)	0
(0, 20)	20
(25,0)	25
$\left(\frac{50}{3}, \frac{40}{3}\right)$	$30 \rightarrow$ Maximum

Thus, maximum value of Z is 30.

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Question 32 Marks

The sum of two positive integers is atmost 5. The difference between two times of second number and first number is at most 4. If first number is x and second number is y, then for maximizing the product of these two numbers, formulate LPP

Answer

The required LPP is
Max. Z = x.y
Subject to:
$\begin{aligned} x+y & \leq 5 \\ 2 y-x & \leq 4 \\ x & \geq 0 \\ y & \geq 0\end{aligned}$

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Question 42 Marks

Draw the feasible region for given inequation system $y \leq 6, x+y \leq 3, x \geq 0, y \geq 0$.

Answer

Thus, feasible region is OAB.

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Question 52 Marks

A firm has to transport at least 1200 packages daily using large vans which carry 200 packages each and small vans which can take 80 packages each. The cost for engaging each large van is ₹ 400 and each small van is ₹ 200. Not more than ₹ 3,000 is to be spent daily on the job and the number of large vans cannot exceed the number of small vans. Formulate this problem as a LPP given that the objective is to minimize cost.

Answer

Let the number of large vans and small vans used for transporting the packages be x and y, respectively. It is given that the cost of engaging each large van is ₹ 400 and each small van is ₹ 200.
Cost of engaging x large van is ₹ 400 and each small van is ₹ 200.
Cost of engaging x large vans = ₹ 400x
Cost of engaging y small vans = ₹ 200y
Let Z be the total cost of engaging x large vans and y small vans.
$\therefore$ Z = ₹ (400x + 200y)
The firm has to transport atleast 1200 packages daily using large vans which carry 200 packages each and small vans which can take 80 packages each.
So, 200x + 80y $\geq$ 1200
Not more than 3000 is to be spent daily on the transport
So, $400 x+200 y \leq 3000$
Also, the number of large vans cannot exceed the number of small vans.
$\therefore$ $x \leq y$
Thus, the linear programming problem of the problem is
Min. Z = ₹ (400x + 200y)
Subject to constraints
$\begin{aligned} 200 x+80 y & \geq 1200 \\ 400 x+200 y & \leq 3000 \\ x & \leq y \\ x & \geq 0, y \geq 0\end{aligned}$

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Question 62 Marks

A small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 24. It takes one hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is ₹100 and that on a bracelet is ₹300. Formulate a LPP for finding how many of each should be produced daily to maximize the profit? It is being given that at least one of each must be produced

Answer

Let x be the number of necklaces manufactured and y be the number of bracelets manufactured. The total number of necklaces and bracelets it can handle is at most 24.
$x+y \leq 24$
The x items takes x hours to manufacture and y items take $\frac{y}{2}$ hours to manufacture and the maximum time available is 16 hours. So,$x+\frac{y}{2} \leq 16$
The profit on one necklace is given as 100 and the profit on one bracelet is given as 300.
Let the profit be Z. To maximize the profit,
Z = 100x + 300y
Therefore, the required is.
Max. Z = 100x + 300y
Subject to constraints,
$\begin{array}{c}x+y \leq 24 \\ x+\frac{y}{2} \leq 16 \\ x, y \geq 1\end{array}$

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Question 72 Marks

For the LPP,
Max. Z = 2x + 3y
The coordinates of the corner points of bounded feasible region are A(3, 3) B(20, 3) C(20,10) D(18, 12) and E(12, 12) Find the maximum value of Z.

Answer

Corner Points	z = 2x + 3y
A(3, 3)	Z = 15
B(20, 3)	Z = 49
C(20,10)	Z = 70
D(18, 12)	Z = 72 $\rightarrow$ Max
E(12, 12)	Z = 60

Thus, maximum value of Z is 72 at D (18, 12).

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Question 82 Marks

A toy company manufactures two types of toys A and B. Demand for toy B is atmost half of that if type A. Write the corresponding constraint if x toys of type A and y toys of type B are manufactured.

Answer

Since, the demand of toys of type B is atmost half of that for toys of type A
$\therefore$ $y \leq \frac{x}{2}$
$2 y-x \leq 0$
$x-2 y \geq 0$

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Question 92 Marks

The feasible region for an LPP is shown in the given Figure. Let Z = 3x - 4y be the objective function. Find the maximum value of Z.

Answer

The feasible region as shown in the figure, has objective function Z = 3x-4y.

Corner points	Corresponding value of Z = 3x-4y
(0, 0)	0
(12, 6)	$12 \leftarrow$ Maximum
(0,4)	$-16 \leftarrow$ Minimum

Hence, the maximum value of Z is 12 at (12,6).

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Question 102 Marks

The feasible solution for a LPP is shown in given figure. Let Z = 3x-4y be the objective function.
Find the minimum value of Z.

Answer

Corner points	Corresponding value of Z = 3x-4y
(0, 0)	0
(0, 0)	$15 \leftarrow$ Maximum
(6, 5)	-2
(6,8)	-14
(4, 10)	-28
(0, 8)	$-32 \leftarrow$ Minimum

Hence, the minimum of Z occurs at (0, 8) and its minimum value is (-32).

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Question 112 Marks

The corner points of the feasible region determined by the system of linear constraints are (0, 0), (0, 40), (20, 40), (60, 20), (60, 0). The objective function is Z = 4x + 3y.
Find the maximum value of Z.

Answer

Corner points	Corresponding value of Z = 4x + 3y
(0, 0)	0
(0, 40)	120
(20, 40)	200
(60, 20)	300 $\leftarrow$ Maximum
(60, 0)	240

The maximum value of Z is 300.

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