SOLUTION: The Whittier Company needs to produce 40 units of Product A tomorrow. They can produce on either machine X or machine Y or both. Each unit of Product A when pro cessed on machine X

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Question 1172947: The Whittier Company needs to produce 40 units of Product A tomorrow. They can produce on either machine X or machine Y or both. Each unit of Product A when pro cessed on machine X takes 30 minutes of time, while a unit processed on machine Y takes 25 minutes. It costs the company $2 per minute and $3 per minute respectively to operate machines X and Y. Tomorrow, machine X has only 10 hours available to produce Product A, while machine Y can be operated as long as desired. Use the graphical method to deter mine how many hours to schedule on each machine to minimize production costs. Draw the isocost lines corresponding to production costs of $2,500, $2,700, $3,000, and $3,500. Which of the isocost lines result in a feasible but suboptimal solution: Which line is both feasible and optimal? Which is not feasible?
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1987)   (Show Source): You can put this solution on YOUR website!
Absolutely! Let's break down this problem step-by-step.
**1. Define Variables**
* Let `x` be the number of units produced on machine X.
* Let `y` be the number of units produced on machine Y.
**2. Formulate Constraints**
* **Production Requirement:** x + y = 40 (The company needs to produce 40 units).
* **Machine X Time Constraint:** 30x ≤ 600 (Machine X has 10 hours = 600 minutes available). This simplifies to x ≤ 20.
* **Non-negativity:** x ≥ 0, y ≥ 0 (You can't produce a negative number of units).
**3. Formulate the Objective Function (Cost Function)**
* Cost = (30 minutes/unit * $2/minute) * x + (25 minutes/unit * $3/minute) * y
* Cost = 60x + 75y
**4. Graph the Constraints**
* **x + y = 40:**
* When x = 0, y = 40.
* When y = 0, x = 40.
* **x = 20:** This is a vertical line at x = 20.
* **x ≥ 0, y ≥ 0:** This limits the solution to the first quadrant.
**5. Find the Feasible Region**
* The feasible region is the area on the graph that satisfies all the constraints. It will be a polygon bounded by the constraint lines.
**6. Draw the Isocost Lines**
* We'll draw lines representing different cost levels:
* $2,500: 60x + 75y = 2500
* $2,700: 60x + 75y = 2700
* $3,000: 60x + 75y = 3000
* $3,500: 60x + 75y = 3500
To graph them, find the x and y intercepts of each line.
* 2500:
* x=0, y = 2500/75 = 33.33
* y=0, x = 2500/60 = 41.67
* 2700:
* x=0, y = 2700/75 = 36
* y=0, x = 2700/60 = 45
* 3000:
* x=0, y = 3000/75 = 40
* y=0, x = 3000/60 = 50
* 3500:
* x=0, y = 3500/75 = 46.67
* y=0, x = 3500/60 = 58.33
**7. Find the Optimal Solution**
* The optimal solution is the point within the feasible region where the isocost line is lowest (minimizing cost).
* The corner points of the feasible region are critical.
* Corner points of the feasible region:
* (20,0)
* (20,20)
* (0,40)
* Test the corner points in the cost function:
* (20,0): 60(20) + 75(0) = $1200
* (20,20): 60(20) + 75(20) = $2700
* (0,40): 60(0) + 75(40) = $3000
* The minimum cost is $1200, which occurs at (20,0).
**8. Analyze Isocost Lines**
* **Feasible and Suboptimal:**
* $2,700: This line intersects the feasible region, but it's not the lowest cost line.
* $3,000: This line also intersects the feasible region, but is not the lowest cost.
* **Feasible and Optimal:**
* The isocost line that goes through the point (20,0) will be the optimal. The cost of 1200 is too low to be one of the isocost lines provided. We can create the 1200 isocost line by calculating the intercept points.
* x=0, y = 1200/75 = 16
* y=0, x = 1200/60 = 20
* **Not Feasible:**
* $3,500: Although the line can be drawn, to reach that cost with the given constraints, the solution would be outside the feasible region when trying to minimize cost.
**Answers**
* **Optimal Solution:** Produce 20 units on machine X and 0 units on machine Y.
* **Feasible but Suboptimal Isocost Lines:** $2,700 and $3,000.
* **Feasible and Optimal Isocost Line:** The isocost line that crosses (20,0) with a cost of $1200.
* **Not Feasible Isocost Line:** $3,500.
Answer by ikleyn(52835)   (Show Source): You can put this solution on YOUR website!
.
The Whittier Company needs to produce 40 units of Product A tomorrow.
They can produce on either machine X or machine Y or both.
Each unit of Product A when pro cessed on machine X takes 30 minutes of time,
while a unit processed on machine Y takes 25 minutes.
It costs the company $2 per minute and $3 per minute respectively to operate machines X and Y.
Tomorrow, machine X has only 10 hours available to produce Product A,
while machine Y can be operated as long as desired.
Use the graphical method to deter mine how many hours to schedule on each machine to minimize production costs.
Draw the isocost lines corresponding to production costs of $2,500, $2,700, $3,000, and $3,500.
Which of the isocost lines result in a feasible but suboptimal solution:
Which line is both feasible and optimal? Which is not feasible?
~~~~~~~~~~~~~~~~~~~~~~~~


        This problem can be easily solved MENTALLY in the head
                using nothing except of common sense. 


Since the cost producing one unit on machine X is $2 against $3 cost of producing on machine Y,
choose machine X to work as long as possible (10 hours).


During this time, 10 hours or 600 minutes, machine X will make 10/0.5 = 20 units of product A.
It will cost 2*600 = 1200 dollars.


Let machine Y makes the rest of the job and produces the rest 40-20 = 20 units of product A.
It will take 25*20 = 500 minutes and will cost 3*500 = 1500 dollars.


The total of 1200 + 1500 = 2700 dollars will be minimum cost to complete the job.

Solved from the very beginning to the very end, using common sense ONLY and without using any other technique.


                The reasons in the post by @CPhill are INCORRECT
              and lead to the incorrect solution and incorrect answer.



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