Consider the Linear programme $$(LP)$$
Max $$4x$$ + $$6y$$
Subject to
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,3x + 2y \le 6 \cr & \,\,\,\,\,\,\,\,\,\,\,2x + 3y \le 6 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x,y \ge 0 \cr} $$

After introducing slack variables $$s$$ and $$t$$, the initial basic feasible solution is represented by the table below (basic variables are $$s=6$$ $$t=6,$$ and the objective function value is $$0$$).

After some simplex iterations, the following table is obtained

From this, one can conclude that

Consider a linear programming problem with two variables and two constraints. The objective function is: Maximize $${x_1} + {x_2}.$$ The corner points of the feasible region are $$(0,0), (0,2), (2,0)$$ and $$(4/3, 4/3).$$

If an additional constraint $${x_1} + {x_2} \le 5$$ is added, the optimal solution is

Consider a linear programming problem with two variables and two constraints. The objective function is: Maximize $${x_1} + {x_2}.$$ The corner points of the feasible region are $$(0,0), (0,2), (2,0)$$ and $$(4/3, 4/3).$$

Let $${y_1}$$ and $${y_2}$$ be the decision variables of the dual and $${v_1}$$ and $${v_2}$$ be the slack variables of the dual of the given linear programming problem. The optimum dual variables are

A company produces two types of toys: $$P$$ and $$Q.$$ Production time of $$Q$$ is twice that of $$P$$ and the company has a maximum of $$2000$$ time units per day. The supply of raw material is just sufficient to produce $$1500$$ toys (of any type) per day. Toy type $$Q$$ requires an electric switch which is available @ $$600$$ pieces per day only. The company makes a profit of Rs.$$3$$ and Rs.$$5$$ on type $$P$$ and $$Q$$ respectively. For maximization of profits, the daily production quantities of $$P$$ and $$Q$$ toys should respectively be