The Janie Gioffre Drapery Company makes three types of draperies at two differen

Question

The Janie Gioffre Drapery Company makes three types of draperies at two different locations. At location I, it can make 10 pairs of deluxe drapes, 20 pairs of better drapes, and 13 pairs of standard drapes per day. At location II, it can make 20 pairs of deluxe, 50 pairs of better, and 6 pairs of standard per day. The company has orders for 1000 pairs of deluxe drapes, 2100 pairs of better drapes, and 600 pairs of standard drapes. If the daily costs are $650 per day at location I and $800 per day at location II, how many days should Janie schedule at each location in order to fill the orders at minimum cost? Find the minimum cost.

Explanation / Answer

Hints
- Location I per day : 10 pairs of deluxe drapes, 20 pairs of better drapes, 13 pairs of standard drapes.
- Location I costs per day is $500.
- Location II per day : 20 pairs of deluxe drapes, 50 pairs of better drapes, 6 pairs of standard drapes.
- Location II costs per day is $800.
- Company wants : 2000 pairs of deluxe drapes, 4200 pairs of better drapes, 1200 pairs of standard drapes.
- How many days to schedule to keep the cost at minimum cost?
- What is the cost?

Building Equations
Let P = Cost of production, Q = days of production at Location I, R = days of production at Location II.
P = 500Q + 800R
Let W = Number of diapers needed, x = number of production of pairs of deluxe drapes, y = .... better..., z = ... standard ....
W = (10x+20y+13z)Q + (20x+50y+6z)R
From the question, W = 2000x + 4200y + 1200z
2000x + 4200y + 1200z = (10Q+20R)x + (20Q+50R)y + (13Q+6R)z

Comparing the coefficient of x :
2000 = 10Q+20R
Comparing the coefficient of y :
4200 = 20Q+50R
Comparing the coefficient of z :
1200 = 13Q+6R

Why I can use the method of comparing coefficient? Because the drapes are not interchangeable. You can't trade deluxe for better or vice versa.

Building the Inequalities to satisfy the demand.
Logical Thinking : You can produce more-than-demanded drapes, but not less. In real life, you pay RM10 for the item of RM8, will you accept when the change is less than RM2?  
2000 <= 10Q + 20R
4200 <= 20Q + 50R
1200 <= 13Q + 6R
Q>=0, R>=0 (It is impossible that the days of production is negative

Solving the problem
Arrange the inequalities in form of y = mx + c, in this case, you can have Q = mR + c, or R = mQ + c. It does not matter which to use because their value are correlated to each other.
Then draw the graph out, find the intersection area. Using the graphical method to find out the value of Q and R in the graph. Then you will know, how many days in location I and location II should be used to keep the cost at minimum, as we have noted it earlier : Q = days of production at Location I, R = days of production at Location II.

After that, substitute the values of Q and R into the production cost equation that we formed earlier :
P = 500Q + 800R

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