Let the factory makes x pieces of item A and y pieces of item B.
Time required by item A (one piece)
cutting = 5 minutes
assembling = 10 minutes
Time required by item B (one piece)
cutting = 8 minutes
assembling = 8 minutes
Total time
cutting = 3 hours & 20 minutes,
assembling = 4 hours
Profit on one piece
item A = Rs. 5, item B = Rs. 6
Thus, our problem is maximized
Z = 5x + 6y
Subject to x \(\ge\) 0, y \(\ge\) 0
5x + 8y\(\le\) 200
10x + 8y \(\le\)240 
From figure, possible points for maximum value of
Z are at (24, 0), (8, 20), (0, 25).
At (24, 0), Z = 120
At (8, 20), Z = 40 + 120 = 160 (Maximum)
At (0, 25), Z = 150

Let x and y units of packet of mixes are purchased from 5 and T respectively. If Z is the total cost, then
Z = 10x + 4y ...(i)
is objective function which we have to minimize
Here constraints are:
4x + y \(\ge\)80..(ii)
2x + y \(\ge\)60.(iii) 
Also, x \(\ge\) 0...(iv)
y \(\ge\)0
On plotting the graph of above constraints or inequalities (ii), (iii), (iv) and (v),
we get shaded region having corner point A, P, B as feasible region. For co-ordinate of P

Point of intersection of
2x + y = 60...(vi)
and 4x + y = 80....(vii)
From (vi) - (vii),
2x + y - 4x - y = 60 - 80
\(\Rightarrow\) -2x = -20
\(\Rightarrow\) x = 10
\(\Rightarrow\) y = 40
\(\because\) Co-ordinate of P = (10, 40)
Now the value of Z is evaluated at corner point the following table:

Since feasible region is unbounded. Therefore we have to draw the graph of the inequality.
10x + 4y < 260 ...(viii)
Since the graph of inequality (viii) does not have any point common.
So the minimum value of Z is 260 at (10, 40). i.e., minimum cost of each bottle is ~ 260 if the company purchases 10 packets of mixes from 5 and 40 packets of mixes from supplier T.

Let x kg and y kg of food X and Y be mixed for the minimum cost of mixture, then LPP is
Minimise,
Z = 6x + 10 y
Subject to :
x + 2y \(\ge \) 10
2x + 2y \(\ge \) 12 \(\Rightarrow\) x + y \(\ge \) 6
3x + y \(\ge \) 8
x, y \(\ge \) 0
Correct graph

Region is unbounded
Rs. 52 i.e., 6x + 10y < 52 or 3x + 5y < 26 has no point common with feasible region.
\(\therefore\) The LPP has optimum solution at (2, 4) and least cost of the mixture = Rs. 52
Value: Balanced diet is essential for healthy body.

Let the investment in bond A be Rs. x and in bond B Rs. y.
Obective function is 
\(Z=\frac { x }{ 10 } +\frac { 9 }{ 100 } y\)
Subject to constraints
x + y \(\le \) 50,000; x \(\ge \) 20,000;
y \(\ge \) 10,000, x\(\ge \)  y(*)

Vertices of feasible region are A, B, C and D.

Return is maximum when Rs. 40,000 are invested in Bond A and Rs. 10,000 in Bond B maximum return is Rs. 4,900.
Since there are more than 3 constraints, student may be given full 6 marks even if reaches upto (*).

Let the mixture contain x kg of food I and y kg of food II.
Getting the objective function as 
Z = 5x + 7y
Getting the constraints
2x + y \(\ge \) 8
x + 2y \(\ge \)10
x, y \(\ge \)  0

Getting the corners of feasible region as A(0, 8), B(2, 4), C(10, 0).
Z_A = 5 \(\times\) 10 + 7 \(\times\) 8
= 56
Z_B = 5 \(\times\)  2 + 7 \(\times\) 4
= 38 (minimum)
Z_c = 5 \(\times\) 10 + 7 \(\times\) 10
= 50
Since 5x + 7y < 38 has no common region with the feasible region.
\(\therefore\) For minimum cost x = 2 kg and y = 4 kg.

Let x be the number of pieces of item A and y be the number of pieces of item B.
The total profit in (Rs) = P = 50x + 28y...(i)
So, we have to maximize P with following constraints :
300x + 150y \(\le \) 15,000..(ii)
(investment constraint)
x + y\(\le \) 80 ...(iii)
(Storage constraint)
x \(\ge \) 0  (non-negative constraint)
y \(\ge \)0  ..........(iv)
300x + 150y = 15,000
\(\Rightarrow\) 2x + y = 100
Solving  2x + y = 100
and  x + y = 80,
we get   x = 20, y = 60

The feasible shaded region (OABC) is shown in graph, then four points O, A, B and C are critical points. Substituting the values of these points in profit equation, we get

The maximum profit of Rs. 2,680 is for 20 pieces of A and 60 pieces of B.

Let the company manufactures sweaters of type A = x, type B = y, daily.
\(\therefore\)  LPP is maximize. P = 200x + 20y s.t.
360x + 120y \(\le \) 72000 
\(\Rightarrow\) 3x + y \(\le \) 300
x + y \(\le \) 300
x - y  \(\le \) 100
\(3x+y=600,\begin{cases} x=0,y=600 \\ y=0,x=200 \end{cases}\)
\(x+y=300,\begin{cases} x=0,y=300 \\ y=0,x=300 \end{cases}\)
\(x-y=100,\begin{cases} x=100,\quad y=0 \\ y=100,\quad x=200 \end{cases}\)
\(\\ x\ge 0\)
\(y\ge 0\)

Getting vertices of feasible region as, O(0, 0), A(100, 0), B(175, 75), C(150, 150) and D(0, 300)
Maximum profit is P = 200(175) + 20(75)
= 35000 + 1500 = Rs. 36500

Given inequations are
\(2 x+4 y \leq 8 \text { or } \ x+2 y \leq 4 \)
\(3 x+y \leq 6, x+y \leq 4, \ x \geq 0, \ y \geq 0 \)
Maximise Z = 2x + 5y on plotting the graph of the inequations we notice shaded portion as feasible solution

Possible points for maximum Z are A(2, 0),\(B\left(\frac{8}{5}, \frac{6}{5}\right)\) C (0,2)

Z is maximum at qo, 2), i.e. x = 0, y = 2, maximum value = 10

Let x and y be nut packages and bolt packages produced each day, respectively.
\(\therefore\)  LPP is Maximize Z = 24x + 18y
Subject to 2x + 3y \(\le \) 10
3x + 2y \(\le \) 10
x, y \(\ge \) 0

Vertices are
\(A\left( 0,\frac { 10 }{ 3 } \right) ,B(2,2)\quad and\quad C\left( \frac { 10 }{ 3 } ,0 \right) \)

Hence, 2 nuts & 2 bolts to be produced to get max. profit of Rs. 84.

Let the manufacturer produces the products A and B be x and y units, respectively.
We construct the following table
\(\begin{array}{c|c|c|c|c} \hline \text { Products } & \begin{array}{c} \text { Produce } \\ \text { (in units) } \end{array} & \begin{array}{c} \text { Time on } \\ \text { Machine I } \\ \text { (in hours) } \end{array} & \begin{array}{c} \text { Time on } \\ \text { Machine II } \\ \text { (in hours) } \end{array} & \begin{array}{c} \text { Profit } \\ \text { (in Rs.) } \end{array} \\ \hline \begin{array}{c} A \\ B \end{array} & \begin{array}{c} x \\ y \end{array} & \begin{array}{c} 3 x \\ 2 y \end{array} & \begin{array}{c} 3 x \\ 7 y \end{array} & \begin{array}{c} 7 x \\ 4 y \end{array} \\ \hline \text { Total } & x+y & 3 x+2 y & 3 x+y & 7 x+4 y \\ \hline \text { Availability } & & 12 & 9 & \\ \hline \end{array}\) 
Here, total profit z = 7x + 4y
i.e. maximise Z = 7x + 4y
subject to the constraints
\(3 x+2 y \leq 12\) 
\(3 x+y \leq 9\) 
and \(x \geq 0, y \geq 0\) 
Now, consider the given inequations as equations
\(3 x+2 y=12\)   ...(i)
3x + y = 9   ...(ii)
Table for line \(3 x+2 y=12 \text { or } y=\frac{12-3 x}{2}\) is
\(\begin{array}{c|c|c} \hline x & 0 & 4 \\ \hline y & 6 & 0 \\ \hline \end{array}\) 
It psses through the points (0, 6) and (4, 0).
On putting (0, 0) in the inequality \(3 x+2 y \leq 12\), we get
\(0+0 \leq 12 \Rightarrow 0 \leq 12\) 
So, the half plane is towards the origin.
Table for line 3x + y = 9 or y = 9 - 3x is
\(\begin{array}{c|c|c} \hline x & 0 & 3 \\ \hline y & 9 & 0 \\ \hline \end{array}\) 
It passes through the points (0, 9) and (3, 0).
On putting (0, 0) in the inequality \(3 x+y \leq 9\), we get
\(0+0 \leq 9 \Rightarrow 0 \leq 9\) 
So, the half plane is towards the origin
Also, \(x \geq 0 \text { and } y \geq 0\) so the region lies in 1st quadrant.
Now, the intersection point oflines (i) and (ii) is
\((3 x+2 y)-(3 x+y)=12-9\) 
\(\Rightarrow y=3\) 
\(\text { and } \quad 3 x=12-2 \times 3\) 
\(\Rightarrow 3 x=12-6\) 
\(\Rightarrow x=2\) 
Thus, the point of intersection is B (2, 3).
The graph of inequations is shown below

Here, we see that OABC is a required feasible region, whose corner points are 0(0, 0), A(3, 0), B(2,3) and C{0, 6).
The values of Z at these corner points are as follows
\(\begin{array}{c|l} \hline \text { Corner points } & z=7 x+4 y \\ \hline O(0,0) & z=0+0=0 \\ A(3,0) & Z=7 \times 3+0=21 \\ B(2,3) & z=7 \times 2+4 \times 3=26 \text { (maximum) } \\ C(0,6) & Z=7 \times 0+4 \times 6=24 \end{array}\) 
Hence, for maximum profit, manufacturer produce 2 units of product A  3 units of product B.