Question

Maximize Z = 2x + 3y
Subject to
$$\begin{gathered} x+y\ge 1 \\ 10x+y\ge 5 \\ x+10y\ge 1 \\ x,y\ge 0 \end{gathered}$$

Answer 1

First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 1, 10x +y = 5, x + 10y = 1, x = 0 and y = 0

Region represented by x + y ≥ 1:
The line x + y = 1 meets the coordinate axes at A(1, 0) and B(0,1) respectively. By joining these points we obtain the line x + y = 1.
Clearly (0,0) does not satisfies the inequation x + y ≥ 1. So,the region in xy plane which does not contain the origin represents the solution set of the inequation x + y ≥ 1.

Region represented by 10x +y ≥ 5:
The line 10x +y = 5 meets the coordinate axes at $C\left(\frac{1}{2},0\right)$ and D(0, 5) respectively. By joining these points we obtain the line
10x +y = 5.Clearly (0,0) does not satisfies the inequation 10x +y ≥ 5. So,the region which does not contains the origin represents the solution set of the inequation 10x +y ≥ 5.

Region represented by x + 10y ≥ 1:
The line x + 10y = 1 meets the coordinate axes at $A\left(1,0\right)$ and $F\left(0,\frac{1}{10}\right)$ respectively. By joining these points we obtain the line
x + 10y = 1.Clearly (0,0) does not satisfies the inequation x + 10y ≥ 1. So,the region which does not contains the origin represents the solution set of the inequation x + 10y ≥ 1.

Region represented by x ≥ 0 and y ≥ 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 feasible region determined by the system of constraints x + y ≥ 1, 10x +y ≥ 5, x + 10y ≥ 1, x ≥ 0, and y ≥ 0, are as follows.

The feasible region is unbounded.Therefore, the maximum value is infinity i.e. the solution is unbounded.

Disclaimer:
The obtained answer is for the given question. Answer in the book is 2.It would be 2 if the question is to minimize Z instead of to maximize Z.