Let the distance covered with speed of 25km/h = x hm
and the distance covered with speed of
40 km/h = y km
The.total distance covered = z km
The L.P.P. of the above problem, therefore, is
Maximize z = x + y
Subject to constraints
4x + 5y \(\le \)200
x/50 + y/40 \(\le\)1
x\(\ge \)0, y\(\ge\)0
Any one value

Let 'x' kg of food I and 'y' kg of food II be mixed we have the table:

Thus LPP problem is as below:
Minimize  Z = 5x + 7y ...(1)
Subject to: \(2x+y\ge 8\) ...(2)
\(x+2y\ge 10\) ...(3)
and \(x\ge 0,y\ge 0\) ...(4)

Draw the lines x + 2y = 10 and 2x + y = 8, x = 0 and The lines x + 2y = 10 and 2x + y = 8 meet at E(2, 4).
First of all, we locate the region represented by (2)-(4).
The shaded region, as shown above, is feasible region.
Applying Corner Point Method, we have:

Hence, the minimum cost = Rs. 38 when 2kg of food I and 4 kg of food II are mixed.
These days people are aware of having balanced, healthy and nutrious diet. For this reason, people get advice from dieticians.

Let 'x' food packets and 'y' food packets be transported from kitchen A to schools P and Q respectively.
Then (60-x-y) food packets will be transported to school R.


Thus we have:
\(x\ge 0,y\ge 0\) and \(60-x-y\ge 0\)
i.e.\(x\ge 0,y\ge 0\) and \(x+y\le 60\)
\(40-x\ge 0,40-y\ge 0,x-y-10\ge 0.\)
Total transportation cost Z is given by:
Z = 5x + 4y + 4(40-x) + 2(40-y) + 5(x+y-10)+3(60-x-y)
= 3x + 4y + 270.
Thus problem reduces to:
Z = 3x + 4y + 270 ...(1)
Subject to: \(x\ge 0,y\ge 0\) ...(2)
\(x+y\le 60\) ...(3)
\(x\le 40\) ..(4)
\(y\le 40\)...(5)
and  \(x+y\ge 10\) ...(6)

The feasible region is as known shaded in the above figure, which is bounded.
Applying Corner Point Method, we have:

Minimum cost is Rs. 300 at the point I (10, 0)
∴ From Kitchen A: 10 packets, 0 packet and 50 packet to Schools P, Q and R respectively.
From Kitchen B: 30 packets, 40 packet and 0 packet to Schools P, Q and R respectively.

Wheat 400 g and rice 200 g at a minimum cost of Rs. 30. We must take balanced healthy diet for good health. Wheat, rice, milk, fruits, nut etc.
We must take balanced healthy diet for good health; Wheat, rice, milk, fruits, nut etc.

Let x bags of rice and y bags of wheat are purchased.
According to given conditions, LPP is,
Maximise Z = 11x + 9y
subject to the constraints 
\(180 x+120 y \leq 1500\)
\(\Rightarrow 3 x+2 y \leq 25\)
\(\text {and } x+y \leq 10 \text {, proceed. }\)
5 bag of rice, 5 bags of wheat maximum profit Rs. 100.

Let x tins of soft drink A and y tins of soft drink Bare produced.
LPP is, Maximise Z = 4x + 3y
subject to the constraints
\(x \geq 0, y \geq 0,4 x+2 y \leq 46,x+3 y \leq 24\)
\(\begin{array}{|c|c|c|c|} \hline & \begin{array}{c} \text { Soft drink } \\ \boldsymbol{A} \end{array} & \begin{array}{c} \text { Soft drink } \\ \boldsymbol{B} \end{array} & \\ \hline \text { Pineapple (litre) } & 4 & 2 & \leq 46 \\ \hline \text { Orange juice (litre) } & 1 & 3 & \leq 24 \\ \hline \text { Profit (in Rs) } & 4 & 3 & \\ \hline \end{array}\)
Sketch the region and find optimum solution.
9 tins of A, 5 tins of B.

Let 'x' units of food 'X' and 'Y' units of food Y be used.
The LLP is:
Minimize  Z = 4x + 3y
Subject to: \(200x+100y\ge 4000\) 
i.e.,\(2x+y\ge 40\)
\(x+2y\ge 50\)
\(40x+40y\ge 1400\) 
i.e., \(x+y\ge 35\)
and \(x\ge 0,y\ge 0\).
The feasible region is shown shaded in the followings figure:

x + 2y = 50 and x + y = 35 meet at G(20, 15) (Solve!)
x + y = 35 and 2x + y = 40 meet at H (5, 30) (Solve!)
Applying Corner Point Method, we have:

Hence, least cost is Rs. 110 when 5 units of food A and 30 units of food B are used.

Let the merchant stock x desktop models and y portable models. Therefore,
x ≥ 0 and y ≥ 0
The cost of a desktop model is Rs. 25000 and of a portable model is Rs. 4000. However, the merchant can invest a maximum of Rs. 70 lakhs.
25000x + 40000y ≤ 7000000
5x + 8y ≤ 1400
The monthly demand of computers will not exceed 250 units.
x + y <= 250`
The profit on a desktop model is Rs 4500 and the profit on a portable model is Rs 5000.
Total profit, Z = 4500x + 5000y
Thus, the mathematical formulation of the given problem is
Maximum Z = 4500x   + 5000 y  ...(1)
subject to the constraints,
\(5 x+8 y \leq 1400 \)
\(x+y \leq 250 \)
\(x, y \geq 0 \)
The feasible region determined by the system of constraints is as follows.

The corner points are A (250, 0), B (200, 50), and C (0, 175).
The values of Z at these corner points are as follows.

The maximum value of Z is 1150000 at (200, 50).
Thus, the merchant should stock 200 desktop models and 50 portable models to get the maximum profit of Rs. 1150000.

Let 'x' and 'y' be the number of gold rings and chains respectively.
We have:
\(x\ge 0\)  ...(1) 
\(y\ge 0\)...(2)
\(x+y\le 24\)...(3) 
\(x+\frac { y }{ 2 } \le 16\) ...(4)
The objective function, or the profit, Z is:
Z = 300x + 190y ..(5)
We have to maximise Z subject to (1)-(4).
For solution set, we draw the lines:
x = 0, y = 0, x + y = 24, 2x + y = 32.
The lines x + y = 24 and 2x + y = 32 meet at E (8,16).

The shaded portion represents the feasible region, which is bounded.
Applying Corner Point Method, we have:

Hence, the maximum profit is Rs. 5,440 and it is obtained when 8 gold rings and 16 chains are manufactured.

Let there be x cakes of first kind and y cakes of second kind. Therefore, x ≥ 0 and y ≥ 0
The given information can be complied in a table as follows.
Subject to: \(200x+100y\le 5000\)
\(\Leftrightarrow 2x+y\le 50\)  ...(1)
\(25x+50y\le 1000\)...(2)
\(\Leftrightarrow x+2y\le 40\)
and  \(x,y\ge 0.\)

 

Total numbers of cakes, Z, that can be made are, Z = x + y
The mathematical formulation of the given problem is
Maximize Z = x + y … (1)
subject to the constraints,
\(2 x+y \leq 50 \)
\(x+2 y \leq 40 \)
\(x, y \geq 0 \)

The feasible region determined by the system of constraints is as follows.
The corner points are A (25, 0), B (20, 10), O (0, 0), and C (0, 20).
The values of Z at these corner points are as follows.

Thus, the maximum numbers of cakes that can be made are 30 (20 of one kind and 10 of the other kind).

Let the young man drives x km and y km at 25kmh and 40 km!h speed respectively, then the LPP is Maximise distance:
Z = x + y
Subject to
2x + 5y \(\le\)100
\(x \geq 0, y \geq 0, \frac{x}{25}+\frac{y}{40} \leq 1\)
x, y \(\ge\)0
Plotting the inequations as graph we notice shaded portion is feasible solution.

Possible points for maximum Z are A(25, 0), \(B\left(\frac{50}{3}, \frac{40}{3}\right)\) and C(0, 20).

Z  is maximum at \( B\left(\frac{50}{3}, \frac{40}{3}\right) \text {, i.e. } x=\frac{50}{3}, y=\frac{40}{3} \text {. }\)
Hence, he must travel \( \frac{50}{3} \mathrm{~km} \)at a speed of  25 km h
\(\text { and } \frac{40}{3} \mathrm{~km} \text { at a speed of } 40 \mathrm{~km} / \mathrm{h} \)for a maximum  distance of 30 km.

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 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 packages of nuts and y packages of bolts be produced each day.
LPP is maximise
P = 17.5x + 7y
Subject of x + 3y \(\le \) 12
3x + y \(\le \) 12
x \(\ge \) 0, y\(\ge \) 0

Vertices of feasible region are A(0, 4), B(3, 3), C(4, 0).
Profit is maximum at B(3, 3) i.e., 3 packages of nuts and 3 packages of bolts

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

Let the number of lamps and shades manufactured be x and y respectively.
\(\therefore\)  LPP is maximise.
Z = 25x + 15y
Subject to 2x + y \(\le \) 12
3x + 2y \(\le \) 20
x \(\ge \) 0, y\(\ge \)  0

Vertices of feasible region are O(0, 0), A (6, 0), B(4, 4) and C (0, 10).
\(\begin{array}{|c|c|c|c|} \hline & \begin{array}{c} \text { Grinding/cutting } \\ \text { machine } \end{array} & \text { Sprayer } & \text { Profit } \\ \hline \text { Pedestal lamps } & 2 \mathrm{hr} & 3 \mathrm{hr} & \text { Rs 25 } \\ \hline \text { Wooden shades } & 1 \mathrm{hr} & 2 \mathrm{hr} & \text {Rs 15} \\ \hline \leq 12 \mathrm{hr} & \leq 20 \mathrm{hr} & \\ \hline \end{array}\)
P(A) = 150, P(B) = 160, P(C) = 150 and For max. profit no. of lamps = 4
no. of shades = 4
\(\begin{array}{|c|c|c|} \hline \text { Points } & Z=25 x+15 y & \text { Values } \\ \hline A(6,0) & 150+0 & 150 \\ \hline B(4,4) & 100+60 & 160 \\ \hline C(0,10) & 0+150 & 150 \\ \hline \end{array}\)
Maximum Profit = Rs. 160.

Let x and y be electronic and manually operated sewing machine purchased respectively.
\(\therefore\) LPP is Maximize.
P = 22x + 18y
Subject to, 
360x + 240y \(\ge \) 5,760
\(\Rightarrow\)  3x + 2y \(\le \) 48
x + y \(\le \) 20
x \(\ge \) 0 , y \(\ge \)  0

Vertices of feasible region are:
A (0, 20), B(8, 12), C(16, 0) & O(0, 0)
P(A) = 360, P(B) = 392. P(C) = 352
\(\therefore\) For Maximum P, Electronic machines = 8, and Manual machines = 12.

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.