AP DSC School Assistant Math Model Test 3

Concept:A linear programming problem's feasible region is the intersection of all constraints. If the feasible region reduces to a single point, the objective function attains only one value, which is both the minimum and the maximum.Explanation:Convert each inequality into an equality to plot the boundaries:$x + y = 6$, $2x + 3y = 3$, $x = 3$, $y = 3$.Consider the direction of each inequality:- $x + y \le 6$: The origin $(0,0)$ satisfies $0 \le 6$, so the region is below the line.- $2x + 3y \ge 3$: The origin does not satisfy $0 \ge 3$, so the region is away from the origin (above the line).- $x \ge 3$ and $y \ge 3$: Regions to the right of $x=3$ and above $y=3$.Intersecting all these conditions gives only one point: $(3,3)$. Check: $3+3=6$ (satisfies first), $2(3)+3(3)=15 \ge 3$ (satisfies second), and $x=3, y=3$ satisfy the last two. No other point can satisfy all constraints.Now evaluate $z = 5x + 7y$ at $(3,3)$:$z = 5(3) + 7(3) = 15 + 21 = 36$.(Note: The original solution incorrectly gave 50; correct calculation is $5\times3=15$, $7\times3=21$, sum = 36.)Since only one point forms the feasible region, the objective function has a single value. In linear programming, that single value is simultaneously the minimum and the maximum.Answer:Option C: either minimum or maximum.