Solution of the following $LP$ problem
Minimize $z = -3x + 2y$
subject to $0 \leq x \leq 4, 1 \leq y \leq 6, x + y \leq 5$ is $.....$

Find the maximum value of $z = 2x + 6y$ subject to the constraints $-x + y \leq 1$,$2x + y \leq 2$,$x \geq 0$,and $y \geq 0$.

If $Z=10x+25y$ subject to $0 \leq x \leq 3, 0 \leq y \leq 3, x+y \leq 5, x \geq 0, y \geq 0$,then $Z$ is maximum at the point:

$A$ diet of a sick person must contain at least $4000$ units of vitamins,$50$ units of proteins,and $1400$ calories. Two foods $A$ and $B$ are available at a cost of ₹ $4$ and ₹ $3$ per unit respectively. If one unit of $A$ contains $200$ units of vitamins,$1$ unit of protein,and $40$ calories,while one unit of food $B$ contains $100$ units of vitamins,$2$ units of protein,and $40$ calories,formulate the problem so that the diet is the cheapest.

The objective function $Z = 4 x_1 + 5 x_2$,subject to $2 x_1 + x_2 \geq 7$,$2 x_1 + 3 x_2 \leq 15$,$x_2 \leq 3$,$x_1, x_2 \geq 0$ has minimum value at the point

$A$ company makes $3$ models of calculators: $A, B$ and $C$ at factory $I$ and factory $II.$ The company has orders for at least $6400$ calculators of model $A, 4000$ calculators of model $B$ and $4800$ calculators of model $C.$ At factory $I, 50$ calculators of model $A, 50$ of model $B$ and $30$ of model $C$ are made every day; at factory $II, 40$ calculators of model $A, 20$ of model $B$ and $40$ of model $C$ are made every day. It costs $Rs. 12000$ and $Rs. 15000$ each day to operate factory $I$ and $II,$ respectively. Find the number of days each factory should operate to minimize the operating costs and still meet the demand.