2. A gold processor has two sources of gold ore, source A and source B. In order to keep his 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 less 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 2 oz. of gold per ton, and ore from source B yields 3 oz. of gold per ton, how many tons of ore from both sources must be processed each day to maximize the amount of gold extracted subject to the above constraints?

Steps

2. Express the objective function and the constraints.
3. Graph the constraints.
4. Find the cornerpoints to the region of feasible solutions.
5. Evaluate the objective function at all the feasible corner points.

In this case they are looking for the number of tons of ore from the two sources which should be processed each day to maximize the amount of gold extracted.

### Step 1. Define the unknowns

Let   x = the number of tons from source A

and   y = the number of tons from source B

### Step 2. Express the objective and the constriants.

The objective is to maximize the amount of the gold yield. Since each ton of ore from source A yields 2oz. of gold and each ton of ore from source B yields 3oz. of gold, the amount of gold recovered will be

2x + 3y

After getting the unknowns and the objective out of the way, everything else in the problem is a constraint. The constraints are the

processing

x + y ≥ 3

cost

20x + 10y ≤ 80

federal regulations

y ≤ 2x

Of course there are also the implied constraints

x ≥ 0   y ≥ 0

We are not going to process an negative number of tons of ore.

### Step 3. Graph the constraints

The implied constraints tell us that the graph will be in the upper right quadrant of the plane.

When we graph the first constraint,

x + y ≥ 3

the x and y intercepts are both 3

Since this is a "greater than" constraint the side of the line where the x's and y's get bigger will be the correct side. This region goes on forever.

In the next constraint,

20x + 10y ≤ 80

the x intercept is 80/20 = 4, and the y intercept is 80/10 = 8. There is enough money to process 4 tons of ore from source A if we don't process any ore from source B, whereas, if we just use ore from source B, there will be enough money to process 8 tons.

Since this is a "less than" constraint, the points on the side where x and y get small should be shaded. The set of points which satisfy both of theses constraints are shaded above.

Finally there is the federal regulations constraint.

y ≤ 2x

In this type of equation, the x and y intercept will both be at (0, 0), so we will need to plot at least one other point. If we let   x = 4   then if   y = 2x   we will have   y = 8.   Plot the point (4, 8) and connect it with (0, 0).

To tell which side of this line is which, notice that the x's are on the large side of the inequality and the y's are on the small side. So we shade the side of the line where the x's are big and the y's are small.

We label the lines with the equations from which they come so that we can

### 4. Find the feasible corner points.

Where we see a   y   we substitute   2x,   and the first equation becomes

x + 2x = 3

3x = 3

x = 1

If   x = 1   and   y = 2x   then   y = 2.   Hence

A = (1, 2)

Again, where we see a   y   we substitute   2x,   and the first equation becomes

20x + 10(2x) = 80

20x + 20x = 80

40x = 80

x = 2

If   x = 2   and   y = 2x   then   y = 4.   Hence

B = (2, 4)

C   is the   x   intercept for   x + y = 3,   so

C = (3, 0)

D   is the   x   intercept for   20x + 10y = 80,   so

D = (4, 0)

### 5. Evaluate the objective function.

At (1, 2) we get 2(1) + 3(2) = 8oz. of gold.

At (2, 4) we get 2(2) + 3(4) = 16oz. of gold.

At (3, 0) we get 2(3) + 3(0) = 6 oz. of gold.

At (4, 0) we get 2(4) + 3(0) = 8oz. of gold.

The maximum yield of gold is 16oz. by processing 2 tons of ore from source A and 4 tons from source B.

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Math 131

Steve Wilson