Consider the following algebraic formulation of a resource-allocation problem with three resources, where the decisions to be made are the levels of three activities (A1, A2, and A3).
Maximize Profit = 20A1 + 40A2 + 30A3
subject to-
Resource 1: 3A1 + 5A2 + 4A3 ? 400 (amount available)
Resource 2: A1 + A2 + A3 ? 100 (amount available)
Resource 3: A1 + 3A2 + 2A3 ? 200 (amount available)
and
A1 ? 0 A2 ? 0 A3 ? 0
A. Identify variables
B. Create an objective function
C. Graph feasible region.
D. Provide optimal solution
Answer:
A) Here, variables are : A1, A2, A3.
B) Let the total profit be $P.
Then, the objective function is : P = 20A1+40A2+30A3.
C) Graph of the given inequalities is :
Here, feasible region is a line whose points are of the form (k,k,100-2k), where k is real number.
[Changing given inequalities as equations and solving them we get the form of points].
D) Putting k = 0 we get,
The point becomes = (0,0,100)
Then, P = 20*0+40*0+30*100
i.e., P = 3000
You may put any real number in place of k that satisfy the condition 
Therefore, optimal solution is = (k,k,100-2k) where k is real number and maximum profit is $3000.
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