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# linear programing

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The Book Emporium buys books from two suppliers: OHaganBooks.com and JungleBooks.com. OHaganBooks.com offers packages of 5 mystery novels and 5 romance novels at \$50 per package, and JungleBooks.com offers packages of 5 mystery novels and 10 romance novels at \$150 per package. The Emporium needs to buy at least 2500 mystery novels and at least 3500 romance novels. For reasons too complicated to explain, at most 75% of the total number of packages must be from OHaganBooks.com. How many packages should the Emporium order from each supplier to satisfy the above requirements at the least possible cost?

Guest Nov 4, 2015

#3
+78741
+9

Minimize the objective function  150x + 50y  where x is the number of packages ordered from Jungle and y is the numer of packages ordered from OHagan

And each package ordered from both companies supplies 5 mystery novels......and we must have at least 2500 of these....so, the constraint becomes :

5x + 5y >= 2500  →  x + y >= 500

And each package from Jungle supplies 10 romance novels and each package from OHagan supplies 5 romance novels.....and we need at least 3500 of these....so we have

10x + 5y >= 3500  →  2x + y >= 700

Further....at least 75% of the the packages must come from OHagan....in other words, out of every 4 ordered, at least 3 must come from OHagan......put another way, we have

3x <=  y

Look at the following graph :   https://www.desmos.com/calculator/6hdrqjgy2o

Since mins and maxes occur only at the corner point of a feasible region, the only point that "works" here occurs at (140, 420)..........

Thus.........  150(140) + 50(420) = \$42000 is the minimum cost by ordering 140 packages from Jungle and 420 packages from OHagan....

CPhill  Nov 4, 2015
edited by CPhill  Nov 4, 2015
edited by CPhill  Nov 4, 2015
edited by CPhill  Nov 4, 2015
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#2
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420 packages from O'Hagan and 140 packages from Jungle would seem to get you 2800 mystery novels and 3500 romances at a total cost of 42000 dollars.

nb.

420/(420+140) = 0.75

Guest Nov 4, 2015
#3
+78741
+9

Minimize the objective function  150x + 50y  where x is the number of packages ordered from Jungle and y is the numer of packages ordered from OHagan

And each package ordered from both companies supplies 5 mystery novels......and we must have at least 2500 of these....so, the constraint becomes :

5x + 5y >= 2500  →  x + y >= 500

And each package from Jungle supplies 10 romance novels and each package from OHagan supplies 5 romance novels.....and we need at least 3500 of these....so we have

10x + 5y >= 3500  →  2x + y >= 700

Further....at least 75% of the the packages must come from OHagan....in other words, out of every 4 ordered, at least 3 must come from OHagan......put another way, we have

3x <=  y

Look at the following graph :   https://www.desmos.com/calculator/6hdrqjgy2o

Since mins and maxes occur only at the corner point of a feasible region, the only point that "works" here occurs at (140, 420)..........

Thus.........  150(140) + 50(420) = \$42000 is the minimum cost by ordering 140 packages from Jungle and 420 packages from OHagan....

CPhill  Nov 4, 2015
edited by CPhill  Nov 4, 2015
edited by CPhill  Nov 4, 2015
edited by CPhill  Nov 4, 2015
#4
+91045
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CPhill has given a top answer here.  Thanks Chris

This is Chris's graph, I have changed it just a bit and if you bounce across to Desmos using my link you will see that I have added a slider for total cost.  You can see that the cheapest orders that meet all requirements is \$72,000

https://www.desmos.com/calculator/kvshzdlail

I have also put a black border around the region of the graph that meets all the requirements.

Melody  Nov 6, 2015

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