Our problem is to minimise and maximise
Z = x + 2y                     ....(i)
Subject to constraints
\(\begin{aligned} & x+2 y \geq 100 \end{aligned}\)        ....(ii)
\(\begin{aligned} 2 x-y \leq 0 \end{aligned}\)            ....(iii)
\(\begin{aligned} 2 x+y \leq 200 \end{aligned}\)        ....(iv)
and \(\begin{aligned} x \geq 0, y \geq 0 \end{aligned}\)     ....(v)
Table for line x + 2y = 100 is

So, the line x + 2y = 100 is passing through the points (0, 50) and (100, 0).
On putting (0, 0) in the inequality x + 2y \(\geq\) 100, we get
0 + 2 \(\times\) 0 \(\geq\) 100
\(\Rightarrow\)     0 \(\geq\) 100 (which is false)
So, the half plane is away from the origin.
Table for line 2x - y = 0 is

So, the line 2x - y = 0 is passing through the points (0, 0) and (10, 20).
On putting (5, 0) in the inequality 2x - y \(\leq\) 0,
we get
2 \(\times\) 5 - 0 \(\leq\) 0
\(\Rightarrow\)     10 \(\leq\) 0 (which is false)
So,the half-plane is towards Y-axis.
Table for line 2x + y = 200 is

So, the line 2x + y = 200 is passing through the points (0, 200) and (100, 0).
On putting (0, 0) in the inequality 2x + y \(\leq\) 200, we get 2 \(\times\) 0 + 0 \(\leq\) 200 \(\Rightarrow\) 0 \(\leq\) 200 (which is true).
So, the half plane is towards the origin.
Also, x, y \(\geq\) 0.
So, the region lies in the lst quadrant.

Clearly, feasible region is ABCDA.
On solving equations 2x - y=0 and x + 2y=100,
we get B(20, 40).
Again, solving the equations 2x - y = 0 and 2x + y = 200, we get C(50, 100).
The corner points of the feasible region are A(0, 50), B(20, 40), C(50, 100) and D(0, 200).
The values of Z at corner points are given below

The maximum value of Z is 400 at D(0, 200) and the minimum value of Z is 100 at all the points on the line segment joining A(0, 50) and B(20, 40).

First of all, let us graph the feasible region of the system of inequalities (2) to (5). The feasible region (shaded). Observe that the feasible region is unbounded.
We now evaluate Z at the corner points.

From this table, we find that – 300 is the smallest value of Z at the corner point (6, 0). Can we say that minimum value of Z is – 300? Note that if the region would have been bounded, this smallest value of Z is the minimum value of Z (Theorem 2). But here we see that the feasible region is unbounded. Therefore, – 300 may or may not be the minimum value of Z. To decide this issue, we graph the inequality
– 50x + 20y < – 300 (see Step 3(ii) of corner Point Method.)
i.e., – 5x + 2y < – 30
and check whether the resulting open half plane has points in common with feasible region or not. If it has common points, then –300 will not be the minimum value of Z.
Otherwise, –300 will be the minimum value of Z. 
It has common points. Therefore, Z = –50 x + 20 y has no minimum value subject to the given constraints.
In the above example, can you say whether z = – 50 x + 20 y has the maximum value 100 at (0, 5)? For this, check whether the graph of – 50 x + 20 y > 100 has points in common with the feasible region.

The feasible region determined by the constraints, 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0, and y ≥ 0, is as follows.


The corner points of the feasible region are A (6, 0) and B (0, 3).
The values of Z at these corner points are as follows.

It can be seen that the value of Z at points A and B is same. If we take any other point such as (2, 2) on line x + 2y = 6, then Z = 6
Thus, the minimum value of Z occurs for more than 2 points.
Therefore, the value of Z is minimum at every point on the line, x + 2y = 6

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 \(\left( \frac { 20 }{ 19 } ,\frac { 45 }{ 19 } \right) \).

Given, Z = -3x + 4y
Subject to the constraints
x + 2y \( \leq\) 8; 3x + 2y \( \leq\) 12 and x \(\geq\)0, y \(\geq\) 0
Now, considering the inequations as equations, we get
  x + 2y = 8                  ...(i)
3x + 2y = 12                ...(ii)
Table for line x + 2y = 8 is

On putting (0, 0) in the inequality x + 2y \( \leq\) 8
0 \( \leq\) 8  (which is true)
So, half plane is towards the origin.
Table for line 3x + 2y = 12

On putting (0, 0) in the inequality 3x + 2y \( \leq\)12
0 \( \leq\) 12 (which is true)
So, half plane is towards the origin.
Also, x \(\geq\) 0 and y \(\geq\) 0, so the feasible region lies in the Ist quadrant.
The point of intersection of Eqs. (i) and (ii) is (2, 3).
The graphical representation of the above system of inequations is given below.

Hence, Z = -12 is minimum at (4, 0).

The system of constraints is:
\(x+y\le 4\) ..(1)
and \(x\ge 0,y\ge 0.\) ...(2)
The shaded region in the following figure is the feasible region determined by the system of constraints (1)-(2)
It is observed that the feasible region OAB is bounded.
Thus we use Corner Point Method to determine the maximum value of Z, where:
Z = 3x + 4y...(3)

The co-ordinates of O, A and B are (0, 0), (4, 0) and (0, 4) respectively.
We evaluate Z at each corner point.

Hence, \(Z_{ max }=16\) at the point (0, 4)

Let x and y be the number of items M and N respectively.
Total profit on the production = Rs. (600 x + 400 y)
Mathematical formulation of the given problem is as follows:
Maximise Z = 600 x + 400 y
subject to the constraints:
x + 2y ≤ 12 (constraint on Machine I) ... (1)
2x + y ≤ 12 (constraint on Machine II) ... (2)
\(x+\frac{5}{4} y \geq 5 \) (constraint on Machine III)  ...(3)
x ≥ 0, y ≥ 0 ... (4)

Let us draw the graph of constraints (1) to (4). ABCDE is the feasible region (shaded) as shown in Figure determined by the constraints (1) to (4). Observe that the feasible region is bounded, coordinates of the corner points A, B, C, D and E are (5, 0) (6, 0), (4, 4), (0, 6) and (0, 4) respectively.
Let us evaluate Z = 600 x + 400 y at these corner points.

We see that the point (4, 4) is giving the maximum value of Z. Hence, the manufacturer has to produce 4 units of each item to get the maximum profit of Rs. 4000.

Maximize P = 3x + 4y + 5z subject to:
\(​​​​​​​2x+3y \le 8,2y+5z\le 10,3x+2y+4z\le 15,x\ge 0,y\ge 0,z\ge 0.\)

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

Let x kg of food I and y kg of food II are mixed.
LPP is, Minimise Z = 50 x + 70 y
subject to the constraints

\(x \geq 0, y \geq 0,2 x+y \geq 8, x+2 y \geq 10\)
On plotting the graph of inequations we notice shaded portion is optimum solution.
Possible points for minimum Z are A( 10, 0), B(2, 4),C(0, 8).
\(\begin{array}{|c|c|c|} \hline \text { Points } & Z=50 x+70 y & \text { Values } \\ \hline A(10,0) & 500+0 & 500 \\ \hline B(2,4) & 100+280 & 380 \\ \hline C(0,8) & 0+560 & 560 \\ \hline \end{array}\)
As region is unbounded, we draw the solution region for 50x+ 70y < 380 or 5x + 7y < 38 and notice no point of solution region (dotted) lies in the feasible region,
Hence, we notice Z is minimum for B(2, 4), i.e. x = 2, y = 4. Hence, 2 kg of food 1 and 4 kg of food 11 must be mixed for a minimum cost of Rs. 380.

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.

Construct the following table of values of objective function

(i) (d): Minimum value of Z is -48 which occurs at (0,8).
(ii) (a): Maximumvalue of Z is 20, which occurs at (5,0).
(iii) (b): Maximum of Z - Minimum of Z
= 20 - (-48) = 20 + 48 = 68
(iv) (c): The corner points of the feasible region are O(0,0), A(3, 0), B(3, 2), C(2, 3), D(0, 3).
(v) (d)

(i) (b): Let B(x, y) be the point of intersection of the given lines
2x + 5y = 100 ....(i)
and \(\frac{x}{25}+\frac{y}{40}=1 \Rightarrow 8 x+5 y=20\)...(ii)
Solving (i) and (ii), we get
\(x=\frac{50}{3}, y=\frac{40}{3}\)
ஃ The point of intersection \(B(x, y)=\left(\frac{50}{3}, \frac{40}{3}\right)\)
(ii) (d): The corner points of the feasible region shown in the given graph are
\((0,0), A(25,0), B\left(\frac{50}{3}, \frac{40}{3}\right), C(0,20)\)
(iii) (a): Here Z = x + y

Thus, max Z = 30 occurs at point \(\left(\frac{50}{3}, \frac{40}{3}\right)\)
(iv) (b): 

(v) (c):

(i) (c): When we solve an L.P.P. graphically, the optimal (or optimum) value of the objective function is attained at corner points of the feasible region.
(ii) (d): From the graph of 3x + 4y < 12 it is clear that it contains the origin but not the points on the line 3x + 4y = 12.

(iii) (d): Maximum of objective function occurs at corner points.

(iv) (d): Value of Z = px + qy at (15, 15) = 15p + 15q and that at (0, 20) = 20 q. According to given condition, we have
15p + 15q = 20q \(\Rightarrow\) 15p = 5q \(\Rightarrow\) q = 3p
(v) (b): Construct the following table of values of the objective function: