SOLUTION: A major coffee supplier has warehouses in Seattle and San Jose. The coffee supplier receives orders from coffee retailers in Salt Lake City and Reno. The retailer

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Question 1166325: A major coffee supplier has warehouses in Seattle and San Jose.
The coffee supplier receives orders from coffee retailers in Salt Lake City and Reno.
The retailer in Salt Lake City needs 400 pounds of coffee and the retailer in Reno needs 350 pounds of coffee.
The Seattle warehouse has 700 pounds available,
and the warehouse in San Jose has 500 pounds available.
The cost of shipping from Seattle to Salt Lake City is $2.50 per pound,
from Seattle to Reno $3.50 per pound,
from San Jose to Salt Lake City $2 per pound
and from San Jose to Reno $2.50 per pound.
Find the number of pounds to be shipped from each warehouse to each retailer to minimize the cost.


Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
A major coffee supplier has warehouses in Seattle and San Jose.
The coffee supplier receives orders from coffee retailers in Salt Lake City and Reno.
The retailer in Salt Lake City needs 400 pounds of coffee and the retailer in Reno needs 350 pounds of coffee.
The Seattle warehouse has 700 pounds available,
and the warehouse in San Jose has 500 pounds available.
The cost of shipping from Seattle to Salt Lake City is $2.50 per pound,
from Seattle to Reno $3.50 per pound,
from San Jose to Salt Lake City $2 per pound
and from San Jose to Reno $2.50 per pound.
Find the number of pounds to be shipped from each warehouse to each retailer to minimize the cost.
~~~~~~~~~~~~~~~


Let X  be the amount shipped from Seattle  to Salt Lake City, in pounds.

Let Y  be the amount shipped from Seattle  to Reno.


Then  the amount shipped from San Jose to Reno is (350-Y) pounds,

while the amount shipped from San Jose to Salt Lake City is (400-X) pounds.


The constraints are

    0 <= X <= 400                                    (1)

    0 <= Y <= 350                                    (2)

    X + Y <= 700                                     (3)

    (400-X) + (350-Y) <= 500,  or  X + Y >= 250.     (4)


The objective function to minimize is the cost of shipping

    F = 2.50X + 3.50*Y + 2*(400-X) + 2.50*(350-Y) = 0.5*X + Y + const.


The constant term in expresssion (5) does not matter for the solution, so we can neglect it
                                                                       (taking it equal to zero for the simplicity).


The feasible domain is shown in the figure below.


    


    Plot x = 400 (red), y = 350 (green), X + Y = 700 (blue), X + Y = 250 (purple)



In the plot, you see a rectangle in QI formed by the red and the green lines

and cut by two sloped lines, blue and purple.


The feasibility domain is the part of this rectangle concluded between the sloped lines.



      On the plot, you see also the fifth color line sloped.

      It is the track of the objective function on the coordinate plane.



We should move this line parallel to itself until it will take its LOWEST possible position, still touching the feasibility domain.


Obviously, it will happen at the point (X,Y) with 

    Y = 0,   which due to constraint  X + Y >= 250  gives for X the value of  X = 250.


Thus the optimal solution is to ship


    250           pounds of coffee from Seattle to Salt Lake City,

      0           pounds of coffee from Seattle to Reno,

    400-250 = 150 pounds of coffee from San Jose to Salt Lake City,

    350-  0 = 350 pounds of coffee from San Jose to Reno.

Solved.




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