Chapter 3.2, Problem 6P

Explanation of Solution

Given data:

The farmer Jane owns 45 acres of land and planning to plant with wheat or corn.

On planting wheat, she yields $200 profit and corn yields $300 profit.

Given table:

	Wheat	Corn
Labor	3 workers	2 workers
Fertilizer	2 tons	4 tons

Consider $x_1$ be the acres of land planted with wheat and $x_2$ be the acres of land planted with corn.

Objective function:

${\text{Maximize z=200x}}_{\text{1}}{\text{+300x}}_{\text{2}}$

Considering the constraints,

Constraint 1: Total acres of land used

Constraint 2: Maximum number of workers to be used is 100

Constraint 3: Maximum tons of fertilizers are 120 tons.

Expressing the constraint 1 in terms of $x_1$and $x_2$:

$x_1+x_2=45$

Expressing the constraint 2 in terms of $x_1$and  $x_2$:

$3x_1+2x_2\le 100$

Expressing the constraint 3 in terms of $x_1$, $x_2$and $x_3$ :

$2x_1+4x_2\le 120$

Therefore, the mathematical model of given LP is,

$\text{Maximize}{\text{z=200x}}_{\text{1}}{\text{+300x}}_{\text{2}}$

Subject to the constraints,

$\begin{array}{l}{\text{x}}_{\text{1}}{\text{+x}}_{\text{2}}=45\\ {\text{3x}}_1{\text{+2x}}_{\text{2}}\le 100\\ 2x_1+4x_2\le 120\\ {\text{x}}_1,x_2\ge 0(\text{Sign restriction)}\end{array}$

Converting the inequality constraint without adding any variable:

$\begin{array}{l}{\text{x}}_{\text{1}}{\text{+x}}_{\text{2}}=45\\ {\text{3x}}_1{\text{+2x}}_{\text{2}}\le 100\\ 2x_1+4x_2\le 120\end{array}$

The coordinate points for the constraint $x_1+x_2=45$ is,

If $x_1=0$ then $x_2=45$, The points are $(0,45)$

If $x_2=0$ then $x_1=45$, The points are $(45,0)$

The coordinate points for the constraint $3x_1+2x_2=100$ is,

If $x_1=0$ then $2x_2=100$, The points are $(0,50)$

If $x_2=0$ then $3x_1=100$, The points are $(33.33,0)$

The coordinate points for the constraint $2x_1+4x_2=120$ is,

If $x_1=0$ then $4x_2=120$, The points are $(0,30)$

If $x_2=0$ then $2x_1=120$, The points are $(60,0)$

Therefore, the coordinate point for the constraint $x_1=0$ is $(3,0)$

Graph:

From the above graph, it is known that the vertices of the feasible region lies in the points $\text{A(0,30) and C(33}\text{.33,0)}$.

Calculating the value of the objective function to find the end points:

$\begin{array}{l}6x_1+4x_2=200\\ 2x_1+4x_2=120\\ ---------\\ 4x_1=80\end{array}$

Therefore, the value of $x_1=20$

Substituting the value of $x_1=20$ in the equation $3x_1+2x_2=100$

$\begin{array}{l}3x_1+\end{array}$