The graphical solution set of the system of in-equations $x+y \leq 70, x+2y \leq 100, 2x+y \leq 120, x \geq 0, y \geq 0$ is given by:

There are two types of fertilisers $F_{1}$ and $F_{2}$. $F_{1}$ consists of $10\%$ nitrogen and $6\%$ phosphoric acid,and $F_{2}$ consists of $5\%$ nitrogen and $10\%$ phosphoric acid. After testing the soil conditions,a farmer finds that she needs at least $14\,kg$ of nitrogen and $14\,kg$ of phosphoric acid for her crop. If $F_{1}$ costs $Rs\,6/kg$ and $F_{2}$ costs $Rs\,5/kg$,determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

$A$ manufacturer of electronic circuits has a stock of $200$ resistors,$120$ transistors and $150$ capacitors and is required to produce two types of circuits $A$ and $B$. Type $A$ requires $20$ resistors,$10$ transistors and $10$ capacitors. Type $B$ requires $10$ resistors,$20$ transistors and $30$ capacitors. If the profit on type $A$ circuit is $Rs. 50$ and that on type $B$ circuit is $Rs. 60$,formulate this problem as a $LPP$ so that the manufacturer can maximize his profit.

For the feasible region $OCDBO$ shown in the figure,the maximum value of the objective function $z = 3x + 4y$ is:

The maximum value of $2x + y$ subject to $3x + 5y \leq 26$ and $5x + 3y \leq 30, x \geq 0, y \geq 0$ is

$A$ manufacturer produces two models of bikes: Model $X$ and Model $Y$. Model $X$ takes $6$ man-hours to make per unit,while Model $Y$ takes $10$ man-hours per unit. There is a total of $450$ man-hours available per week. Handling and marketing costs are $Rs. 2000$ and $Rs. 1000$ per unit for Models $X$ and $Y$ respectively. The total funds available for these purposes are $Rs. 80,000$ per week. Profits per unit for Models $X$ and $Y$ are $Rs. 1000$ and $Rs. 500$ respectively. How many bikes of each model should the manufacturer produce so as to yield a maximum profit? Find the maximum profit.