A calculator company produces a scientific calculator

LINEAR PROGRAMMING EXAMPLES

• A calculator company produces a scientific calculator and a graphing calculator. Long-term projections indicate an expected demand of at least 100 scientific and 80 graphing calculators each day. Because of limitations on production capacity, no more than 200 scientific and 170 graphing calculators can be made daily. To satisfy a shipping contract, a total of at least 200 calculators much be shipped each day.

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 y: number of graphing 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: 100 < x < 200 80 < y < 170 y > –x + 200

The feasibility region graphs is below, 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".

:06-2011 All Rights Reserved

• You need to buy some filing cabinets. You know that Cabinet X costs $10 per unit, requires six square feet of floor space, and holds eight cubic feet of files. Cabinet Y costs $20 per unit, requires eight square feet of floor space, and holds twelve cubic feet of files. You have been given $140 for this purchase, though you don't have to spend that much. The office has room for no more than 72 square feet of cabinets. How many of which model should you buy, in order to maximize storage volume?

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 y: number of model Y 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 space: 6x + 8y < 72, or y < –( 3/4 )x + 9 volume: V = 8x + 12y

This system (along with the first two constraints) graphs is:

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.

• At a certain refinery, the refining process requires the production of at least two gallons of gasoline for each gallon of fuel oil. To meet the anticipated demands of winter, at least three million gallons of fuel oil a day will need to be produced. The demand for gasoline, on the other hand, is not more than 6.4 million gallons a day.

If gasoline is selling for $1.90 per gallon and fuel oil sells for $1.50/gal, how much of each should be produced in order to maximize revenue?

The question asks for the number of gallons which should be produced, so I should let my variables stand for "gallons produced".

x: gallons of gasoline produced y: gallons of fuel oil produced

Since this is a "real world" problem, I know that I can't have negative production levels, so the variables can't be negative. This gives me my first two constraints: namely, x > 0 and y > 0.

Since I have to have at least two gallons of gas for every gallon of oil, then x > 2y.

For graphing, of course, I'll use the more manageable form "y < ( 1/2 )x".

The winter demand says that y > 3,000,000; note that this constraint eliminates the need for the "y > 0" constraint. The gas demand says that x < 6,400,000.

I need to maximize revenue R, so the optimization equation is R = 1.9x + 1.5y. Then the model for this word problem is as follows:

R = 1.9x + 1.5y, subject to: x > 0 x < 6,400,000 Copyright © Elizabeth S06-2011 All Rights Reserved y > 3,000,000 y < ( 1/2 )x

Using a scale that counts by millions (so "y = 3" on the graph means "y is three million"), the above system graphs as follows:

Taking a closer look, I can see the feasibility region a little better:

When you test the corner points at (6.4m, 3.2m), (6.4m, 3m), and (6m, 3m), you should get a maximal solution of R = $16.96m at (x, y) = (6.4m, 3.2m).

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