Constraint A: $2 X_{1}+2 X_{2} \leq 14$ Constraint B: $4 X_{1}+10 X_{2} \leq 40$
All variables are required to be non-negative. The corner points of the feasible region for this constraint set (assuming the first number in parenthesis is $X_{1}$ and the second number in parenthesis is $X_{2}$ ) is
A) $(0,0),(0,7),(7,0)$
B) $(0,0),(0,4),(7,0)$
C) $(0,0),(0,4),(10,0)$
D) $(0,0),(0,4),(5,2),(10,0)$
E) $(0,0),(0,4),(5,2),(7,0)$

Question

Quizplus · Accepted Answer

The corner points are found by solving the system of inequalities and considering the non-negativity constraints. Solving the equations $2X_1 + 2X_2 = 14$ and $4X_1 + 10X_2 = 40$ gives the intersection point (5,2). The other points are found by setting each variable to zero in turn, giving (0,0), (0,4), and (7,0).

Constraint A 2X1+2X2≤142 X_{1}+2 X_{2} \leq 142X1​+2X2​≤14 Constraint B 4X1+10X2≤404 X_{1}+10 X_{2} \leq 404X1​+10X2​≤40

Constraint A $2 X_{1}+2 X_{2} \leq 14$ Constraint B $4 X_{1}+10 X_{2} \leq 40$