Smart board lesson 3
Linear Programming Word Problems
1. Baking a tray of corn muffins takes 4 cups of milk and 3 cups of wheat flour. Baking a tray of bran muffins takes 2 cups of milk and 3 cups of wheat flour. A baker has 16 cups of m ilk and 15 cups of wheat flour. He makes $3 profit per tray of corn muffins and $2 profit per tray of bran muffins. How many trays of each type of muffin should the baker make to maximize his profits?
Since the problems is asking for the number trays of each muffin type my variables will stand for the number of ounces of each:
x: number of trays of Corn Muffins
y: number of trays of Bran Muffins
| |Corn Muffin trays, x |Bran Muffin trays, y |Total |
|Milk |4x |2y |16 |
|Flour |3x |3y |15 |
|Profit |3x |2y |3x + 2y |
Constraints:
4x + 2y < 16
3x + 3y < 15
x > 0
y > 0
The optimization equation will be the profit relation P= 3x + 2y.
When you test the vertices, you should get a maximum profit of $13 when the baker makes 3 trays of corn muffins and 2 trays of bran muffins.
2. In order to ensure optimal health (and thus accurate test results), a lab technician needs to feed the rabbits a daily diet containing a minimum of 24 grams (g) of fat, 36 g of carbohydrates, and 4 g of protein. But the rabbits should be fed no more than five ounces of food a day. Rather than order rabbit food that is custom-blended, it is cheaper to order Food X and Food Y, and blend them for an optimal mix. Food X contains 8 g of fat, 12 g of carbohydrates, and 2 g of protein per ounce, and costs $0.20 per ounce. Food Y contains 12 g of fat, 12 g of carbohydrates, and 1 g of protein per ounce, at a cost of $0.30 per ounce. What is the optimal blend?
Since the exercise is asking for the number of ounces of each food required for the optimal daily blend, my variables will stand for the number of ounces of each:
x: number of ounces of Food X
y: number of ounces of Food Y
Since I can't use negative amounts of either food, the first two constrains are the usual ones: x > 0 and y > 0. The other constraints come from the grams of fat, carbohydrates, and protein per ounce:
fat: 8x + 12y > 24
carbs: 12x + 12y > 36
protein: 2x + 1y > 4
Also, the maximum weight of the food is five ounces, so:
x + y < 5
The optimization equation will be the cost relation C = 0.2x + 0.3y, but this time I'll be finding the minimum value, not the maximum.
After rearranging the inequalities, the system graphs as:
[pic]
(Note: One of the lines above is irrelevant to the system. Can you tell which one?)
When you test the vertices at (0, 4), (0, 5), (3, 0), (5, 0), and (1, 2), you should get a minimum cost of sixty cents per daily serving, using three ounces of Food X only.
3. The Widget Manufacturing Company makes two products, Alpha Widgets and Beta Widgets. Each Alpha Widget gives a profit of $3, while each Beta Widget earns $7. The company must manufacture at least one Alpha Widget per day to satisfy its customers, but no more than five because of the number of employees needed to create the item. Also, the number of Beta Widgets produced cannot exceed six per day. As a further constraint, Alpha Widgets cannot exceed the number of Beta Widgets. How many of each Widget should the company produce in order to obtain the maximum profit?
The question asks for the optimal number widgets to produce, so my variables will stand for that:
x: number of Alpha Widgets
y: number of Beta Widgets
Constraints:
For the alpha widgets, the company must manufacture at least one Alpha Widget per day to satisfy its customers, but no more than five:
1 < x < 5
The number of Beta Widgets produced cannot exceed six per day, but they can't produce negative numbers of widgets either:
0 < y < 6
Alpha Widgets cannot exceed the number of Beta Widgets
x < y
The optimization equation will be the profit relation P = 3x + 7y.
When you test the vertices points at (1, 6), (5, 6), (5, 5), and (1, 1), you should obtain the maximum value of P = 57 at (5, 6). So, there is a maximum profit of $57 when the company produces 5 alpha widgets and 6 beta widgets each day.
4. 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 –x + 200
The feasibility region graphs as:
[pic]
When you test the vertices 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".
5. 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 as:
[pic]
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.
6. Katie Stein is a sales representative whose territory is Arizona and New Mexico. Her daily travel expenses average $120 in Arizona and $100 in New Mexico. She receives an annual travel allowance of $18,000. She must spend at least 50 days in Arizona and 60 days in New Mexico. If sales average $3000 in Arizona and $2800 in New Mexico, how many days should she spend in each state to maximize sales?
x = # of days in Arizona
y = # of days in New Mexico
Expenses: 120x + 100y < 18000
Number of days: x ≥ 50 y ≥ 60
Objective Function: S = 3000x + 2800y (sales)
The vertices are (50, 60), (50, 120), and (100, 60).
Sales: S = 3000x + 2800y You need to maximize sales.
(50, 60) S = 3000(50) + 2800(60) = 318,000
(50, 120) S = 3000(50) + 2800(120) = 486,000
(100, 60) S = 3000(100) + 2800(60) = 468,000
Therefore, Katie Stein should spend 50 days in Arizona and 120 days in New Mexico.
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