Vocab
2 Variable System
3 Variable System
Linear Programming
Word Problems
100

the solution to a system with 2 variables 

ordered pair 

100

solve this system using substitution: 

x=5

2x + 4y = 22

(5, 3)

100

solve:

y = x + z + 5

z = -3y - 3

2x - y = -4

(-2, 0, -3)

100

Which value would be the maximum?

a) 18

b) 24

c)7

b) 24

100

is the following problem asking to find a maximum or minimum?

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?

Maximize

200

the shaded region in a linear programming problem

feasible region 

200

solve this system using elimination:

x + 3y = 9

2x + 3y = 15

(6, 1)

200

solve:

-2y + 5z = -3

y = -5x - 4z - 5

x = 4z + 4

(0, -1, -1)

200

what quadrant do we typically work in when solving linear programming problems? 

1st quadrant

200

define the variables: 

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.

x = number of scientific calculators produced 

y = number of graphing calculators produced 

300

what is the solution to a system with 3 variables called

ordered triple

300

solve this system using substitution: 

-x + 4y = 6

3x - 2y = 2

(2, 2)

300

solve:

4x + 4y + z = 24

2x - 4y + z = 0

5x - 4y - 5z = 12

(4, 2, 0)

300

given the objective function and vertices, find the minimum:

f(x,y) = 3x + 2y

(4, 0)      (2, - 1)    (-5, 9)   

(-5, 9)

300

write the objective function: 

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?


f(x, y) = -2x + 5y

400

equation that is used to find maximum and minimum values

objective function 

400

solve using elimination: 

5x + y = 9

10x - 7y = -18

(1, 4)

400

solve:

-3x - y - 3z = -8

-5x + 3y + 6z = -4

-6x - 4y + z = -20

(2, 2, 0)

400

The value of 27 matches with which vertex:

f(x, y) = 5x - y

(4, 7)   (6, 3)   (10, 2)   (5, -3)

(6, 3)

400

write the constraints:

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.

x > 100   (all of these should also be equal to)

y > 80

x < 200

y < 170

x + y > 200 

500

the values the maximize or minimize an objective function

vertices 

500

solve using elimination: 

3x - 2y = 2

5x - 5y = 10

(-2, -4)

500

solve:

-6x - 2y + 2z = -8

3x - 2y - 4z = 8

6x - 2y - 6z = -18

no solution

500

is (3, 4) a solution for the following feasible region, EXPLAIN:

x > 0

y > 0 

2x + 4y < 8

no, (3,4) is not a solution

500

The objective function is maximized at the vertex (100, 170) when using the objective function f(x, y) = -2x + 5y. Write a conclusion statement: what is the maximum profit and where does it occur?

x = number of scientific calculators sold 

y = number of graphing calculators sold. 

The company needs to sell 100 scientific calculators and 170 scientific calculators to achieve a maximum profit of $650. 

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