If $z = ax + by$ where $a, b > 0$ subject to the constraints $x \leq 2, y \leq 2, x + y \geq 3, x \geq 0, y \geq 0$ has a minimum value at $(2, 1)$ only,then...

If the difference between the maximum and minimum values of the objective function $z = 7x - 8y$ subject to the constraints $x + y \leqslant 20$,$y \geqslant 5$,$x, y \geqslant 0$ is $5k + 200$,then the value of $k$ is

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 $.....$

The maximum value of $z=50x+15y$ subject to the constraints $x+y \leq 60$,$5x+y \leq 100$,$x \geq 0$,$y \geq 0$ is 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.

$A$ cottage industry manufactures pedestal lamps and wooden shades,each requiring the use of a grinding/cutting machine and a sprayer. It takes $2 \text{ hours}$ on the grinding/cutting machine and $3 \text{ hours}$ on the sprayer to manufacture a pedestal lamp. It takes $1 \text{ hour}$ on the grinding/cutting machine and $2 \text{ hours}$ on the sprayer to manufacture a shade. On any day,the sprayer is available for at most $20 \text{ hours}$ and the grinding/cutting machine for at most $12 \text{ hours}$. The profit from the sale of a lamp is $Rs. 5$ and that from a shade is $Rs. 3$. Assuming that the manufacturer can sell all the lamps and shades that he produces,how should he schedule his daily production in order to maximise his profit?