Q#1:

First, find the feasible region and draw the line X1 + 2X2 = 0
by letting Z=0 in the objective function. Then, we can observe that
the optimal solution is achieved at the vertex of (X1=7, X2=2/3),
which is the solution of the system of equations:
        X1       = 7
        X1 + 3X2 = 9
(You need to solve this system of equations).
Thus, the optimal objective value is
        Z = 7 + 2(2/3) = 25/3.

In summary, the optimal solution of the LP is
        X1 = 7 and X2 = 2/3,
and the optimal objective value is 25/3.


Q#2:

(a)
Given the decision variables X1 and X2, we verify the LP formulation.

We will maximize the profit = price - cost.
Since "the prices of both Product 1 and 2 will be $18 per unit" and
"producing each unit of the two products will cost $1 per machine hour",
we maximize
        Z = (18 - 10 - 2)X1 + (18 - 5 - 3)X2
          = 6X1 + 10X2.

First, we have the restrictions of
        X1 >= 0 and X2 >= 0,
since the units of the products to be produced can not be negative.

From the information given in the 2nd table, total milling machine hours needed is
 		10X1 + 5X2, 
and total lathe hours needed is
		2X1 + 3X2.
Because of the machine capacity limits given in the 1st table, we have 
(1) Milling machine capacity constraint, i.e., 10X1 + 5X2 <= 500 and
(2) Lathe capacity constraint, i.e., 2X1 + 3X2 <= 150.

Since "the sales potential for Product 2 will not exceed 30 units per week", we have
(3) Product 2 sales constraint, i.e., X2 <= 30.
There is no sales constraint on Product 1.

These are all the constraints, and the LP problem has been formulated correctly.


(b)
As in Q#1, first, find the feasible region and draw the line
6X1 + 10X2 = 0 by letting Z=0 in the objective function.
Then, we can observe that the optimal solution is achieved at the vertex
of (X1=30, X2=30), which is the solution of the system of equations:
        2X1 + 3X2 = 150
               X2 = 30
(You need to solve this system of equations).
Thus, the optimal objective value is
        Z = 6(30) + 10(30) = 480.

In summary, the optimal solution of the LP is
        X1 = 30 and X2 = 30,
and the optimal objective value is 480.