.
The restrictions are
63*B + 7*V >= 252 (1) (to transport all 252 students)
2*B + 1*V <= 22 (2) (chaperones)
B >= 0, (3)
V >= 0. (4)
The objective function to minimize is
F(B,V) = 1200*B + 120*V. (5) (transportation cost)
The feasible domain is shown in the figure below
It is the triangle domain in the Quadrant I restricted by the straight lines (inequalities)
63*B + 7*V >= 252 (1) (red line)
2*B + 1*V <= 22 (2) (green line)
V >= 0 (4)
So, it is the area ABOVE the red sloped line and BELOW the green line
lines 63*B + 7*V = 252 (red) and 2*B + 1*V = 22 (green)
The critical points are
P1 = (2,18) (intersection of lines 1) and (2)) F(2,18) = 1200*2 + 120*18 = 4560.
P2 = (4,0) (B-intercept for the red line (1)) F(4,0) = 1200*4 + 120*0 = 4800.
P3 = (11,0) (B-intercept for the green line (2)) F(0,22) = 1200*11 + 120*0 = 13200.
Calculate the function F(B,V) in each of the points P1, P2 and P3.
Then select the point where the function F(B,V) is minimal.
The Linear Programming Method states that this point is the solution: it provides minimal transportation cost.
So, in our case the optimal solution is 2 buses and 18 vans.
The transportation cost is 4560 dollars in this case.