Question

If \( x + y \leq 55 \) and \( x + y \geq 10 \), with \( x \geq 0 \), \( y \geq 0 \), then the minimum value of the objective function \( z = 7x + 3y \) is: \hfill \break

Hint:

For a linear programming problem to have a minimum or maximum value, the feasible region must be bounded within a closed area.

Answer 1

Step 1: Understanding the constraints. The given constraints are: \[ x + y \leq 55, \] \[ x + y \geq 10, \] \[ x \geq 0, \quad y \geq 0. \] These constraints define a feasible region in the first quadrant where the values of \( x \) and \( y \) lie. 

Step 2: Identifying the feasible region. The inequalities define a strip in the first quadrant between the lines: - \( x + y = 10 \) (lower boundary). - \( x + y = 55 \) (upper boundary). However, for a solution to exist, the feasible region should be a bounded region where an optimal solution can be determined. 

Step 3: Checking feasibility for optimization. Since the given constraints do not form a closed bounded region (it extends infinitely), there is no minimum bound for the function \( z = 7x + 3y \). Thus, the solution region is not feasible for determining the minimum value. 

Conclusion: Since the feasible region does not bound the function properly, the minimum value cannot be determined.