Question

A steel plant has two sources of ore, source A and source B. In order to keep the plant running, at least three tons of ore must be processed each day. Ore from source A costs $20 per ton to process, and ore from source B costs $10 per ton to process. Costs must be kept to no more than $80 per day. Moreover, Federal Regulations require that the amount of ore from source B cannot exceed twice the amount of ore from source A. If ore from source A yields 300 lbs of steel per ton, and ore from source B yields 400 lbs of steel per ton, how many tons of ore from each source should be processed each day to maximize the amount of steel produced?
A) 1 ton from source A, 2 tons from source B
B) 2 tons from source A, 4 tons from source B
C) 3 tons from source A, 0 tons from source B
D) 4 tons from source A, 0 tons from source B

Answer 1

The correct answer for the question that is being presented above is this one: "C) 3 tons from source A, 0 tons from source B." The tons of ore from each source should be processed each day to maximize the amount of steel produced is that C) 3 tons from source A, 0 tons from source B 

Answer 2

The answer for this problem is B) 2 tons from source A, 4 tons from source B.

To solve an optimization problem, you can use linear programming or simply algebra. Based on the statements given, you can create three equations for the constraints, using ore A as x and ore B as y.

x + y >= 3 (at least three tons of ore must be processed each day)
20x + 10y <= 80 (costs must be kept to no more than $80 per day)
y/x <= 2 (ore from source B cannot exceed twice the amount of ore from source A)

The function to be optimized in the problem is for the amount of steel produced: 300x + 400y

This gives 2 solutions at points (2, 4) and (1, 2). Now, test which solution MAXIMIZES the production of ore:

@ (2,4): 300*2 + 400*4 = 2200 lbs ore
@(1,2): 300 + 400*2 = 1100 lbs ore

So B is the answer.