Answer to Quiz2

1.A	2. A	3. C	4. D	5. B


6. The feasible region is the area of (0,0), (6,0), (5,2), (2,5), (0,6)

7. The optimal point is n(5,2), objective function value is 6*(5)+4*(2) = 38.

8. The profit will increase because the constraint of resource A is active.

9. The profit won't change because the constraint on resource B is inactive.

10. The profit/unit of product 1 should be at least 8. You can get the result 
by compare the objective value  at the extreme point with that of (6,0),
 	
		6p1 >= 2p1+5*4 = 2p1+20     then   p1 >= 5   [(2,5)], 
   		6p1 >= 5p1+2*4 = 5p1+8	    then   p1 >= 8   [(5,2)],
		6p1 >= 0p1+6*4 = 24	    then   p1 >= 4   [(0,6)]

If  (6,0) is optimal, all of the above should hold, so we get the profit of product 1 
should be at least 8.

Or, you can consider the problem this way: if the coefficient of the objective fuction 
changes,  the slope of the objective line will change. If you want (6,0), which 
is the intersection of 	x2=0  and  2x1+x2 =12,
to be optimal,   the slope of the new objective function should be at least as straight 
as the 2x1+x2. So we have    -p1/4<= -2/1, that is, p1>=8.

1. A	2. A	3. D	4. D	5. B

6. The feasible region is the area of (0,0), (6,0), (5,2), (2,5), (0,6)

7. The optimal point is m(2,5), objective function value is 3*(2)+4*(5) = 26.

8. The profit will remain the same because the constraint of resource A is inactive.

9. The profit will increase because the constraint on resource B is active.

10. The profit/unit of product 2 should be at least 6. You can get the result 
by compare the objective value  at the extreme point with that of (0,6),
 	
		6p2 >= 3*2+5p2 = 6 + 5p2    then      p2 >= 6   [(2,5)], 
   		6p2 >= 3*5 +2p2= 15+2p2     then   p2 >= 3.75   [(5,2)],
		6p2 >= 3*6+0p2= 24	    then   p2 >= 3      [(6,0)]

If  (6,0) is optimal, all of the above should hold, so we get the profit of product 2 
should be at least 6.

Or, you can consider the problem this way: if the coefficient of the objective fuction 
changes, the slope of the objective line will change. If you want (0,6), which is the 
intersection of  	x1=0  and  x1+2x2 =12,
to be optimal, the slope of the new objective function should be at least as straight as 
 x1+2x2. So we have    -3/p2<= -1/2, p1>=6.

1. A	2. A	3. D	4. D	5.  C