An oil company owns two refineries.  The daily production limits and operating
costs for each refinery are given in the table below.  An order is received for
1540 barrels of high-grade oil, 1650 barrels of medium-grade oil, and 2860
barrels of low-grade oil.  How many days should each refinery operate so that
the order can be filled at the least cost?

	                 Refinery 1	Refinery 2
High-grade oil (barrels)	110	220
Medium-grade oil (barrels)	220	110
Low-grade oil (barrels)	        330	220
Operating cost (barrels)	14,000	8,000

		x = number of days refinery 1 should operate.

		y = number of days refinery 2 should operate.

Constraints:



Objective (money intake) Function: P = 14000x + 8000y

Find all the intercepts and points of intersections of all 6 lines --
three slanted boundary lines, 1 vertical y-axis boundary line, whose
equation is x = 0, and 1 horizontal x-axis boudary line whose equation
is y = 0.

{The equation of the line parallel to one of the axes is an equation in 
terms of the other letter, the letter of the other axis.]



which when simplified are



You can find all these points by all the methods you've learned in
algebra using the substitution or the addition method. I found more 
of the points than I needed to in order to get the answer to this 
problem.  I only needed to find the lower corner points, since the
optimum value is always at one of the corner points. 

We'll substitute these corner points of the bottom vertices of an
infinitely-large polygon, which are marked with red dots below: 



The feasible region is the infinite area on or above all 3 slanted lines 
and the x-axis, and on or to the right of the y-axis.

The objective function is always the money coming in or the money going out.  In
this case it's going ouit. It's the function C = 14000x + 8000y.  We evaluate it
at all the 4 red points, the only ones we needed to find the coordinates for.

Corner  |       C=     
point   | 14000x+8000y=
--------|--------------
(0,15)  |   $120000
 (4,7)  |    112000
 (6,4)  |    115000
(14,0)  |    196000  

So the winner is the point (4,7) with a cost of only $112000.

So the best plan is to operate refinery 1 for 4 hours and operate
refinery 2 for 7 hours. 

x = 4 days refinery refinery 1 should operate.

y = 7 days refinery 2 should operate.

Least Cost:   $112,000

Edwin