Two oil refineries each produce three grades of petroleum, A , B and C . At each refinery, the various grades of petroleum are produced in a single operation so that they are in fixed proportions. One operation at Refinery 1 produces 2 units of A , 1 unit of B and 5 units of C . One operation at Refinery 2 produces 5 units of A , 1 unit of B and 2 units of C . Refinery 1 charges $350 for one operation, and Refinery 2 charges $200 for one operation. A consumer needs at least 540 units of A , 180 units of B and 420 units of C . How should the orders be placed if the consumer’s needs are to be met most economically? Each grade of oil, A , B and C , gives an inequality in x and y . Plot these inequalities on a graph. Enter the coordinates of the corners of the feasible region (the feasible basic solutions). Enter them in increasing order of their x -coordinate. For example, if one feasible basic solution is x=1,y=2 ; another is x=5,y=0 and a third is x=2,y=3 , you would enter (1,2), (2,3), (5,0)

We can solve this problem by translating it into a linear programming problem and then graphing the inequalities to determine the feasible region.

### Step 1: Define the variables
Let:
- $ x $ = the number of operations at Refinery 1.
- $ y $ = the number of operations at Refinery 2.

### Step 2: Constraints based on petroleum requirements
We know that:
- Refinery 1 produces $ 2 $ units of $ A $, $ 1 $ unit of $ B $, and $ 5 $ units of $ C $ per operation.
- Refinery 2 produces $ 5 $ units of $ A $, $ 1 $ unit of $ B $, and $ 2 $ units of $ C $ per operation.
- The consumer needs at least $ 540 $ units of $ A $, $ 180 $ units of $ B $, and $ 420 $ units of $ C $.

This gives us the following inequalities:

1. For $ A $: 
   \[
   2x + 5y \geq 540
   \]

2. For $ B $: 
   \[
   x + y \geq 180
   \]

3. For $ C $: 
   \[
   5x + 2y \geq 420
   \]

### Step 3: Objective function (minimizing cost)
The total cost is:
\[
\text{Cost} = 350x + 200y
\]
We need to minimize this cost.

### Step 4: Solve the system of inequalities
We now solve the inequalities for the boundary points of the feasible region. Each inequality will form a straight line, and the region that satisfies all three constraints will be our feasible region.

#### Inequality 1: $ 2x + 5y \geq 540 $
Solve for $ y $:
\[
y \geq \frac{540 - 2x}{5}
\]

#### Inequality 2: $ x + y \geq 180 $
Solve for $ y $:
\[
y \geq 180 - x
\]

#### Inequality 3: $ 5x + 2y \geq 420 $
Solve for $ y $:
\[
y \geq \frac{420 - 5x}{2}
\]

### Step 5: Plotting and finding the corner points
The three lines are:
1. $ y = \frac{540 - 2x}{5} $
2. $ y = 180 - x $
3. $ y = \frac{420 - 5x}{2} $

Next, we find the intersection points of these lines, which are the corner points of the feasible region.

- To find the intersection of $ y = \frac{540 - 2x}{5} $ and $ y = 180 - x $, set them equal:
  \[
  \frac{540 - 2x}{5} = 180 - x
  \]
  Solving this gives $ x = 60 $, $ y = 120 $.

- To find the intersection of $ y = 180 - x $ and $ y = \frac{420 - 5x}{2} $, set them equal:
  \[
  180 - x = \frac{420 - 5x}{2}
  \]
  Solving this gives $ x = 120 $, $ y = 60 $.

- To find the intersection of $ y = \frac{540 - 2x}{5} $ and $ y = \frac{420 - 5x}{2} $, set them equal:
  \[
  \frac{540 - 2x}{5} = \frac{420 - 5x}{2}
  \]
  Solving this gives $ x = 180 $, $ y = 0 $.

Thus, the coordinates of the corners of the feasible region are:
\[
(60, 120), (120, 60), (180, 0)
\]

### Step 6: Conclusion
The corners of the feasible region, in increasing order of their $ x $-coordinate, are:
\[
(60, 120), (120, 60), (180, 0)
\]

Would you like to see the graph or have more details about how the corner points were calculated?

---

#### 5 Related Questions:
1. How do we determine which corner point gives the minimum cost?
2. How would the solution change if the costs at the refineries were different?
3. What happens if the consumer's petroleum needs increase?
4. How would we handle constraints if an additional refinery is introduced?
5. What if the proportions of petroleum grades produced at each refinery change?

#### Tip:
In linear programming, the minimum or maximum value of the objective function always occurs at one of the vertices (corner points) of the feasible region.

Two oil refineries each produce three grades of petroleum, A, B and C. At each refinery, the various grades of petroleum are produced in a single operation so that they are in fixed proportions. One operation at Refinery 1 produces 2 units of A, 1 unit of B and 5 units of C. One operation at Refinery 2 produces 5 units of A, 1 unit of B and 2 units of C. Refinery 1 charges $350 for one operation, and Refinery 2 charges $200 for one operation. A consumer needs at least 540 units of A, 180 units of B and 420 units of C. How should the orders be placed if the consumer’s needs are to be met most economically? Each grade of oil, A, B and C, gives an inequality in x and y. Plot these inequalities on a graph. Enter the coordinates of the corners of the feasible region (the feasible basic solutions). Enter them in increasing order of their x-coordinate. For example, if one feasible basic solution is x=1,y=2; another is x=5,y=0 and a third is x=2,y=3, you would enter (1,2), (2,3), (5,0).

Learn how to minimize costs in petroleum production using linear programming. Solve inequalities for three grades of petroleum from two refineries, plot the feasible region, and find the most economical solution. Ideal for Grades 10-12, this problem covers optimization techniques and graphing inequalities.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation