Linear Programming: Word Problems (page 3 of 5) Sections: Optimizing linear systems, Setting up word problems
If each scientific calculator sold results in a $2 loss, but each graphing calculator produces a $5 profit, how many of each type should be made daily to maximize net profits? The question asks for the optimal number of calculators, so my variables will stand for that: x:
number of scientific calculators produced Since they can't produce negative numbers of calculators, I have the two constraints, x > 0 and y > 0. But in this case, I can ignore these constraints, because I already have that x > 100 and y > 80. The exercise also gives maximums: x < 200 and y < 170. The minimum shipping requirement gives me x + y > 200; in other words, y > –x + 200. The profit relation will be my optimization equation: P = –2x + 5y. So the entire system is: P = –2x + 5y,
subject to: The feasibility region graphs as: Copyright © Elizabeth Stapel 20062011 All Rights Reserved When you test the corner points at (100, 170), (200, 170), (200, 80), (120, 80), and (100, 100), you should obtain the maximum value of P = 650 at (x, y) = (100, 170). That is, the solution is "100 scientific calculators and 170 graphing calculators".
The question ask for the number of cabinets I need to buy, so my variables will stand for that: x:
number of model X cabinets purchased Naturally, x > 0 and y > 0. I have to consider costs and floor space (the "footprint" of each unit), while maximizing the storage volume, so costs and floor space will be my constraints, while volume will be my optimization equation. cost: 10x
+ 20y < 140, or
y <
–( ^{1}/_{2} )x + 7
This system (along with the first two constraints) graphs as: When you test the corner points at (8, 3), (0, 7), and (12, 0), you should obtain a maximal volume of 100 cubic feet by buying eight of model X and three of model Y. << Previous Top  1  2  3  4  5  Return to Index Next >>



Copyright © 20062014 Elizabeth Stapel  About  Terms of Use 




