Linear Programming

Let the manufacturer produce x packages of nuts and y packages of bolts. Therefore,

x ≥ 0 and y ≥ 0

The given information can be compiled in a table as follows.

The profit on a package of nuts is Rs 17.50 and on a package of bolts is Rs 7. Therefore, the constraints a

(i) Let the number of rackets and the number of bats to be made be x and y respectively.

The machine time is not available for more than 42 hours.

$\therefore 1.5x+3y\le 42....\left(1\right)$

The craftsman’s time is not available for more than 24 hours.

$\therefore 3x+y\le 24 ......\left(2\right)$

The factory is to work at full capacity. Therefore,

$\begin{array}{l}1.5x + 3y = 42\hfill \end{array}$

$\begin{array}{l}3x + y = 24\hfill \end{array}$

On solving these equati

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.

$\begin{array}{l}\therefore 200x+100y\le 5000\\ \Rightarrow 2x+y\le 50\\ 25x+50y\le 1000\\ \Rightarrow x+2y\le 40\end{array}$

Total numbers of cakes, Z, that can be made are, $Z= x + y$

The mathematical formulation of the given problem is

Maximize&nb

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

$3x + 4y \ge 8$

$5x + 2y \ge 11$

Total cost, Z, of purchasing

Maximise $z=x+y$ , subject to $x-y\le -1,-x+y\le 0,x,y\ge 0$

The corresponding equation of the given inequalities are

$x-y=1$ $\frac{x}{-1}+\frac{y}{1}=1$

$-x+y=0$ $\Rightarrow$ $x=y$

$x,y=0$ $x,y=0$

The graph of the given inequalities is shown.

There is no common point in the two shaded region. Thus, there is no feasible region.

$\therefore$ Z has no maximum value.

Maximise $z=-x+2y$ , subject to the constraints

$x\ge 3,x+y\ge 5,x+2y\ge 6,y\ge 0.$

The corresponding equation of the given inequalities are

$x=3$ $x=3$

$x+y=5$ $\Rightarrow \frac{x}{5}+\frac{y}{5}=1$

$x+2y=6$ $\frac{x}{6}+\frac{y}{3}=1$

$y=0$ $y=0$

The graph of the given inequalities is shown

The feasible region unbounded.

The values of Z at corner points A (6,0), B (4,1), and C (3,2) are as follows

As the feasible regio

Minimize and Maximise $z=x+2y$

Subject to $x+2y\ge 100,2x-y\le 0,2x+y\le 200,x,y\ge 0$

The corresponding equation of the given inequalities are

$x+2y=100$ $\Rightarrow \frac{x}{100}+\frac{y}{50}=1$

$2x-y=0$ $2x=y$

$2x+y=200$ $\frac{x}{100}+\frac{y}{200}=1$

$x,y\ge 0$ $x,y=0$

The graph of the inequalities is shown below.

The shaded bounded region ABCD is the feasible region with the corner points.

A (0,50), B, (20,40), C (50,100), D (0,200)

The values o

Minimize and Maximise $z=5x+10y$

Subject to $x+2y\le 120,x+y\ge 60,x-2y\ge 0,x,y\ge 0$

The corresponding equation of the given inequalities are

$\begin{array}{l}x+2y=120\\ x+y=60\\ x-2y=0\\ x,y=0\end{array}$

$\Rightarrow \frac{x}{120}+\frac{y}{60}=120$

$\begin{array}{l}\frac{x}{60}+\frac{y}{60}=1\\ x=2y\\ x,y=0\end{array}$

The graph of the inequalities in shown.

The shaded founded region ABCD is the feasible region with the corner points $A\left(60,0\right),B\left(120,0\right),C\left(60,30\right)\&D\left(40,20\right)$

The value of Z a these corner points are

The minimum value of Z is 300 at (60,0)

Minimize $z=x+2y$

Subject to $2x+y\ge 3,x+2y\ge 6,x,y\ge 0$

The corresponding equation of the given inequalities are

$\begin{array}{l}2x+y=3\\ x+2y=6\\ x,y\ge 0\end{array}$

$\Rightarrow \frac{x}{\frac{3}{2}}+\frac{y}{3}=1$

$\frac{x}{6}+\frac{y}{3}=1$

$x,y\ge 0$

The feasible region is unbounded the corner point are A (6,0), B (0,3)

The value of Z at these corner points are follows.

Since, the feasible region is unbounded, a graph of $x+2y<6$ is drawn.

Also since there is no

Maximum $z=3x+2y$

Subject to $x+2y\le 10,3x+y=\le 15,x,y\ge 0$

The corresponding equation of the given inequalities are :

$\begin{array}{l}x+2y=10\\ 3x+y=15\\ x,y=0\end{array}$

$\Rightarrow \frac{x}{10}+\frac{y}{5}=1$

$\begin{array}{l}\frac{x}{5}+\frac{y}{5}=1\\ x,y=0\end{array}$

The graph of the given inequalities

The shaded bounded region OABC in the feasible region with the corner points

$O(0,0),A(5,0),B(4,3),C(0,5)$

The value of Z at these points are given below.

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