Question

Solve the following Linear Programming Problem graphically. Maximise profit \[{\text{Z}} = {\text{Rs }}\left( {80{\text{x}} + 120{\text{y}}} \right)\]
Subject to constraints are
 \[
  9{\text{x}} + 12{\text{y}} \leqslant 180 \\
  1{\text{x}} + 3{\text{y}} \leqslant {\text{30}} \\
  {\text{x}},{\text{y}} \geqslant {\text{0}} \\
 \]

Answer 1

Hint: The linear programming is to find the maximum or minimum value of ${\text{z}}$ by converting the inequalities to equalities according to the given constraints. The intersecting point of the two equations is considered as corner points.

Complete step-by-step answer:
The given data in the question are,
 \[{\text{Z}} = {\text{Rs }}\left( {80{\text{x}} + 120{\text{y}}} \right)\] is maximized under the following constraints,
 \[
  9{\text{x}} + 12{\text{y}} \leqslant 180 \\
  1{\text{x}} + 3{\text{y}} \leqslant {\text{30}} \\
  {\text{x}},{\text{y}} \geqslant {\text{0}} \\
 \]

By converting the inequalities and equalities we get,
 \[
  9{\text{x}} + 12{\text{y}} = 180\,......................(1) \\
  1{\text{x}} + 3{\text{y}} = {\text{30 }}..........................{\text{(2)}} \\

 \]
Step 1
While substituting ${\text{x}} = 0$ in the equation $(1)$ we get,
 ${\text{12y}} = 180$
By solving we get,
 \[{\text{y}} = \dfrac{{180}}{{12}}\]
By dividing the above we get,
 ${\text{y}} = 15$
From the equation $(1)$ , while substituting ${\text{x}} = 0$ we get the points as $(0,15)$
While substituting ${\text{y}} = 0$ in the equation $(1)$ we get,
 ${\text{9x}} = 180$
By solving we get,
 \[{\text{x}} = \dfrac{{180}}{9}\]
By dividing the above we get,
 ${\text{x}} = 20$
From the equation $(1)$ , while substituting ${\text{y}} = 0$ we get the points as $(20,0)$
While substituting ${\text{x}} = 0$ in the equation $(2)$ we get,
 ${\text{3y}} = 30$
By solving we get,
 \[{\text{y}} = \dfrac{{30}}{3}\]
By dividing the above we get,
 ${\text{y}} = 10$
From the equation $(2)$ , while substituting ${\text{x}} = 0$ we get the points as $(0,10)$
While substituting ${\text{y}} = 0$ in the equation $(2)$ we get,
 ${\text{x}} = 30$
From the equation $(2)$ , while substituting ${\text{y}} = 0$ we get the points as $(30,0)$

Step 2
By plotting all the above points in a graph we get the graph as,
               

The two lines intersect at $(12,6)$ and the other corner points of the region are $(0,10),(0,0)\,{\text{and }}(20,0)$ .

 Step 3
To find the maximum profit of ${\text{z}}$ , we have to find all values of ${\text{z}}$ in the corner points,
 \[{\text{Z}} = {\text{Rs }}\left( {80{\text{x}} + 120{\text{y}}} \right)\]
When we substitute $(0,10)$ in the above equation we get,
 \[{\text{Z}} = 0 + 120(10)\]
By solving the above we get,
 \[{\text{Z}} = 1200.....................({\text{i)}}\]
When we substitute $(0,0)$ we get,
 \[{\text{Z}} = 0\,.........................({\text{ii}})\]
When we substitute $(20,0)$ we get,
 \[{\text{Z}} = 80(20) + 0\]
By solving we get,
 \[{\text{Z}} = 1600\,..................({\text{iii}})\]
When we substitute $(12,6)$ we get,
 \[{\text{Z}} = 80(12) + 120(6)\]
By simplifying we get,
 \[{\text{Z}} = 960 + 720\]
By solving we get,
 \[{\text{Z}} = 1680\,..................({\text{iv)}}\]
The maximum value of $({\text{i),(ii),(iii) and (iv)}}$ is \[({\text{iv)}}\] .
Hence, \[{\text{Z}}\] is maximum at $(12,6)$ and the maximum value is $1680$ .
Therefore, the maximum profit is ${\text{Rs}}.1680$

Note: There are three types of linear programming. We have to solve the problem according to the question. In the constraints given we have to note that the value of x and y should be greater than or equal to $0$