The minimum value of $t = 7x + 3y$ subject to constraints $x + y < 5$,$x + y < 10$,$x > 0$,$y > 0$ is . . . . . .

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:

The difference between the maximum and minimum values of the objective function $Z = 3x + 5y$, subject to the constraints $x + 3y \leq 60$, $x + y \geq 10$, $x - y \leq 0$, and $x, y \geq 0$, is

The shaded part of the given figure indicates the feasible region. Then the constraints are

The shaded area in the figure below is the solution set for a certain linear programming problem. Then the linear constraints are given by:

$A$ merchant plans to sell two types of personal computers - a desktop model and a portable model that will cost $Rs.\,25000$ and $Rs.\,40000$ respectively. He estimates that the total monthly demand of computers will not exceed $250$ units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than $Rs.\,70$ lakhs and if his profit on the desktop model is $Rs.\,4500$ and on portable model is $Rs.\,5000$.