The feasible region determined by the constraints, x + y ≤ 4, x ≥ 0, y ≥ 0, is as follows.
The corner points of the feasible region are O (0, 0), A (4, 0), and B (0, 4). The values of Z at these points are as follows.

Therefore, the maximum value of Z is 16 at the point B (0, 4).

200 at (50,0)

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' kg bran and 'y' kg of rice be required.
Minimize: Z = 5x + 4y subject to:
\(x\ge 0,y\ge 0,80x+100y\ge 88\) and \(40x+30y\ge 56\)
i.e. \(x\ge 0,\quad y\ge 0,20x+25y\ge 22\)
and \(20x+15y\ge 28\)

The feasible region (shaded) is unbounded.
Let us evaluate Z at the corner points:

Hence, the minimum cost is Rs. 7 when \(\frac { 14 }{ 10 } \) kg of bran is used.

The system constraints is:
\(x-2y\le 2\)....(1)
\(3x+2y\le 12\) ..(2)
\(-3x+2y\le 3\) ..(3)
and \(x\ge 0,y\ge 0\) ..(4)

The line x - 2y and 3x + 2y = 12 meet at H \(\left( \frac { 7 }{ 2 } ,\frac { 3 }{ 4 } \right) \).
The lines -3x + 2y = 3 and 3x + 2y = 12 meet at G \(\left( \frac { 3 }{ 2 } ,\frac { 15 }{ 4 } \right) \)
The shaded portion in the above figure is the feasible region, which is bounded.
Applying Corner Point Method, we are to determine the maximum and minimum values of Z, where Z = 5x + 2y.

Hence, Z_min = 0 at (0,0) and Z_max = 19 at \(\left( \frac { 7 }{ 2 } ,\frac { 3 }{ 4 } \right) \)

Let x and y be the number of dolls of type A and B respectively that are produced per week.
The given problem can be formulated as follows.
Maximize z = 12x + 16y … (1)
subject to the constraints,
\(x+y\le 1200\) ..(1)
\(y\le \frac { x }{ 2 } \Leftrightarrow x-2y\ge 0\) ..(2)
\(x\le 3y+600\Leftrightarrow x-3y\le 600\) ...(3)
and \(x,y\ge 0\) ....(4)
The feasible region determined by the system of constraints is as follows.

The corner points are A (600, 0), B (1050, 150), and C (800, 400).
The values of z at these corner points are as follows.

The maximum value of z is 16000 at (800, 400).
Thus, 800 and 400 dolls of type A and type B should be produced respectively to get the maximum profit of Rs. 16000.

Let x and y litres of oil be supplied from A to the petrol pumps, D and E. Then, (7000 − x − y) will be supplied from A to petrol pump F.
The requirement at petrol pump D is 4500 L. Since x L are transported from depot A, the remaining (4500 −x) L will be transported from petrol pump B.
Similarly, (3000 − y) L and 3500 − (7000 − x − y) = (x + y − 3500) L will be transported from depot B to petrol pump E and F respectively.
The given problem can be represented diagrammatically as follows.
Minimize: \(C=\frac { 3x }{ 10 } +\frac { y }{ 10 } +3950\)
Subject to:   \(4500-x\ge 0\Leftrightarrow x\le 4500\) ...(1)
\(3000-y\ge 0\Leftrightarrow y\le 3000\)....(2)
\(x+y-3500\ge 0\Leftrightarrow x+y\ge 3500\) ...(3)
\(7000-(x+y)\ge 0\Leftrightarrow x+y\le 7000\) ..(4)
and  \(x,y\ge 0\) ....(5)
Cost of transporting \(10 \mathrm{~L} \text { of petrol }=\operatorname{Re} 1\)
Cost of transporting \( 1 \mathrm{~L} \text { of petrol }=\mathrm{Rs} \frac{1}{10}\)
Therefore, total transportation cost is given by,
\(z =\frac{7}{10} \times x+\frac{6}{10} y+\frac{3}{10}(7000-x-y)+\frac{3}{10}(4500-x)+\frac{4}{10}(3000-y)+\frac{2}{10}(x+y-3500)\)
\(=0.3 x+0.1 y+3950 \)
The problem can be formulated as follows.
Minimize z = 0.3x + 0.1y + 3950 … (1)
subject to the constraints,
\(x+y \leq 7000 \)
\(x \leq 4500 \)
\(y \leq 3000 \)
\(x+y \geq 3500 \)
\(x, y \geq 0 \)
The feasible region determined by the constraints is as follows.


The corner points of the feasible region are A (3500, 0), B (4500, 0), C (4500, 2500), D (4000, 3000), and E (500, 3000).
The values of z at these corner points are as follows.

The minimum value of z is 4400 at (500, 3000).
Thus, the oil supplied from depot A is 500 L, 3000 L, and 3500 L and from depot B is 4000 L, 0 L, and 0 L to petrol pumps D, E, and F respectively.
The minimum transportation cost is Rs. 4400.

 Let the farmer mix x bags of brand P and y bags of brand Q.
The given information can be compiled in a table as follows.

The given problem can be formulated as follows.
Minimize z = 250x + 200y … (1)
subject to the constraints,
\(3x+1.5y\ge 18\) 
\(2.5x+11.25y\ge 14\) 
\(2x+3y\ge 24\) 
and \(x,y\ge 0\)  
The feasible region determined by the system of constraints is as follows

The corner points of the feasible region are A (18, 0), B (9, 2), C (3, 6), and D (0, 12).
The values of z at these corner points are as follows.

As the feasible region is unbounded, therefore, 1950 may or may not be the minimum value of z.
For this, we draw a graph of the inequality, 250x + 200y < 1950 or 5x + 4y < 39, and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with 5x + 4y < 39
Therefore, the minimum value of z is 1950 at (3, 6).
Thus, 3 bags of brand P and 6 bags of brand Q should be used in the mixture to minimize the cost to Rs. 1950. 

Let the mixture contain x kg of food P and y kg of food Q. Therefore, x ≥ 0 and y ≥ 0
The given information can be compiled in a table as follows.

The mixture must contain at least 8 units of vitamin A and 11 units of vitamin B. Therefore, the constraints are
3x + 4y ≥ 8
5x + 2y ≥ 11
Total cost, Z, of purchasing food is, Z = 60x + 80y
The mathematical formulation of the given problem is
Minimise Z = 60x + 80y … (1)
subject to the constraints,
3x + 4y ≥ 8 … (2)
5x + 2y ≥ 11 … (3)
x, y ≥ 0 … (4)
The feasible region determined by the system of constraints is as follows.
It can be seen that the feasible region is unbounded.
The corner points of the feasible region are and \(A\left(\frac{8}{3}, 0\right), B\left(2, \frac{1}{2}\right) \text { and } C\left(0, \frac{11}{2}\right)\)

As the feasible region is unbounded, therefore, 160 may or may not be the minimum value of Z.
For this, we graph the inequality, 60x + 80y < 160 or 3x + 4y < 8, and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with 3x + 4y < 8
Therefore, the minimum cost of the mixture will be Rs 160 at the line segment joining the points \(\left(\frac{8}{3}, 0\right) \text { and }\left(2, \frac{1}{2}\right)\)

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

The corner points of the feasible region are A (5, 0), B (4, 3), and C (0, 5).
The values of Z at these corner points are as follows.

Therefore, the maximum value of Z is 18 at the point (4, 3).

First of all, let us graph the feasible region of the system of linear inequalities (2) to (5). The feasible region ABCD. Note that the region is bounded. The coordinates of the corner points A, B, C and D are (0, 10), (5, 5), (15,15) and (0, 20) respectively

We now find the minimum and maximum value of Z. From the table, we find that the minimum value of Z is 60 at the point B (5, 5) of the feasible region.
The maximum value of Z on the feasible region occurs at the two corner points C (15, 15) and D (0, 20) and it is 180 in each case.

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' 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.

True.

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 (*).

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.