A producer has 30 and 17 units of labour and capital respectively which he can use to produce two type of goods x and y. To produce one unit of x, 2 units of labour and 3 units of capital are required. Similarly, 3 units of labour and 1 unit of capital is required to produce one unit of y. If x and y are priced at Rs 100 and Rs 120 per unit respectively, how should be producer use his resources to maximize the total revenue? Solve the problem graphically.
Solution
Let \[x_1\] and \[y_1\] units of goods x and y were produced respectively.
Number of units of goods cannot be negative.
Therefore,\[x_1 , y_1 \geq 0\]
To produce one unit of x, 2 units of labour and for one unit of y, 3 units of labour are required \[2 x_1 + 3 y_1 \leq 30\]
To produce one unit of x, 3 units of capital is required and 1 unit of capital is required to produce one unit of y
\[3 x_1 + y_1 \leq 17\]
If x and y are priced at Rs 100 and Rs 120 per unit respectively, Therefore, cost of x_{1} and y_{1} units of goods x and y is Rs 100x_{1} and Rs 120y_{1} respectively.
Total revenue = Z = \[100 x_1 + 120 y_1\] which is to be maximised.
Max Z = \[100 x_1 + 120 y_1\]
\[3 x_1 + y_1 \leq 17\]
First we will convert inequations into equations as follows:
2x_{1} + 3y_{1} = 30, 3x_{1} + y_{1} = 17, x = 0 and y = 0
Region represented by 2x_{1} + 3y_{1} ≤ 30:
The line 2x_{1} + 3y_{1}_{ }= 30 meets the coordinate axes at A(15, 0) and B(0, 10) respectively. By joining these points we obtain the line
2x_{1} + 3y_{1}_{ }= 30. Clearly (0,0) satisfies the 2x_{1} + 3y_{1}_{ }= 30. So, the region which contains the origin represents the solution set of the inequation 2x_{1} + 3y_{1} ≤ 30.
Region represented by 3x_{1} + y_{1} ≤ 17:
The line 3x_{1} + y_{1} = 17 meets the coordinate axes at
3x_{1} + y_{1} = 17. Clearly (0,0) satisfies the inequation 3x_{1} + y_{1} ≤ 17. So,the region which contains the origin represents the solution set of the inequation 3x_{1} + y_{1} ≤ 17.
Region represented by x_{1} ≥ 0 and y_{1} ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The corner points are B(0, 10), E(3, 8) and \[C\left( \frac{17}{3}, 0 \right)\]
The values of Z at these corner points
orner point  Z= \[100 x_1 + 120 y_1\] 
B  1200 
E  1260 
C 
\[\frac{1700}{3}\]

The maximum value of Z is 1260 which is attained at E(3, 8).
Thus, the maximum revenue is Rs 1260 obtained when 3 units of x and 8 units of y were produced.are as follows