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.

We have the data:

Let 'x' chairs and 'y' tables be produced per week.
Then we have:
\(2x+y\le 70\) ...(1)
\(x+y\le 40\)...(2)
\(x+3y\le 90\) ...(3)
\(x\ge 0,y\ge 0\) ....(4)
and  \(Z=30x+60y\) .....(5)

The feasible region (shaded) OAPF is bounded, where O is (0, 0), A is (35, 0), P(24, 22) and F is (0, 30).
Applying Corner Point Method, we have:

Hence, the maximum profit is Rs. 2,040 when 24 chairs and 22 tables are made per week.​​​

Let the man travel 'x' travel 'x' km at 25 km/hr. and 'y' km at 40 km/hr.
Clearly \(x\ge 0\)...(1) and  \(y\ge 0\)    ...(2) 
Since the total money is Rs.100,
∴ \(2x+5y\le 100\)...(3)
Since the total time is 1 hour,
∴ \(\frac { x }{ 25 } +\frac { y }{ 40 } \le \) i.e. \(8x+5y\le 200\)..(4)

The objective function or the distance D, in one hour is:
D = X + Y..(5)
Draw the lines:
x = 0, y = 0, 2x + 5y = 100
and 8x + 5y = 200.
The feasible region (shown shaded) OCEB is bounded,
where O is (0, 0), C is (25, 0), B is (0, 20) and E is \(\left( \frac { 50 }{ 3 } ,\frac { 40 }{ 3 } \right) \).
 \(\left[ ∵Solving2x+5y=100and8x+5y=200;x=\frac { 50 }{ 3 } ,y=\frac { 40 }{ 3 } \right] \)
Applying Corner Point Method, we have:

Hence, maximum distance travelled = 30 km.

The feasible region determined by the system of constraints, 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, and y ≥ 0, are as follows.

The corner points of the feasible region are O (0, 0), A (2, 0), B (0, 3), and C\(\left( \frac { 20 }{ 19 } ,\frac { 45 }{ 19 } \right) \)
The values of Z at these corner points are as follows.

Therefore, the maximum value of Z is \(\frac { 235 }{ 19 } \) at \(\left( \frac { 20 }{ 19 } ,\frac { 45 }{ 19 } \right) \).

Maximize Z = 30x + 60y subject to:
\(​​​​​​​2x+y\le 70,x+y\le 40,x+3y\le 90,x,y\ge 0.\)

Let x hectare of land be allocated to crop X and y hectare to crop Y. Obviously,
x ≥ 0, y ≥ 0.
Profit per hectare on crop X = Rs. 10500
Profit per hectare on crop Y = Rs. 9000
Therefore, total profit = Rs. (10500x + 9000y)


The mathematical formulation of the problem is as follows:
Maximise Z = 10500 x + 9000 y subject to the constraints:
x + y ≤ 50 (constraint related to land) ... (1)
20x + 10y ≤ 800 (constraint related to use of herbicide)
i.e. 2x + y ≤ 80 ... (2)
x ≥ 0, y ≥ 0 (non negative constraint) ... (3)
Let us draw the graph of the system of inequalities (1) to (3). The feasible region OABC is shown (shaded) in the Figure Observe that the feasible region is bounded.
The coordinates of the corner points O, A, B and C are (0, 0), (40, 0), (30, 20) and (0, 50) respectively. Let us evaluate the objective function Z = 10500 x + 9000y at these vertices to find which one gives the maximum profit.

Hence, the society will get the maximum profit of Rs. 4,95,000 by allocating 30 hectares for crop X and 20 hectares for crop Y.

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 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 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.

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