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Chapter 11 - Wisniewski

Essay by   •  February 1, 2017  •  Coursework  •  1,702 Words (7 Pages)  •  1,322 Views

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A toy manufacturer preparing a production schedule for two new toys, trucks and spinning tops, must use the information concerning their manufacturing times, given in the following table.

 

Machine A

Machine B

Finishing

Truck

2 hr

3 hr

5 hr

Spinning Top

1 hr

1 hr

1 hr

The available hours per week per machine are: Machine A, 80 hours; Machine B, 50 hours; Finishing, 70 hours.  If the profits on each truck and spinning top are $7 and $2 respectively, how many of each toy should be made per week in order to maximize profit?  What is the maximum profit?

Decision Variables:

        x = number of trucks produced per week

y = number of spinning tops produced per week

Objective Function:

Maximize profit                7x + 2y

Subject to:

Machine A capacity limit (in hours):         2x + 1y ≤ 80 hours/week

Machine B capacity limit (in hours)        3x + 1y ≤ 50 hours/week

Finishing capacity limit (in hours)        5x + 1y ≤ 70 hours/week

Non-negativity                                x, y ≥ 0

Calculations:

Graphing the constraints:

The x- and y- intercepts for the of the constraints are

Constraint

x-intercept

y-intercept

Machine A capacity

2x + 1y ≤ 80

(40, 0)

(0, 80)

Machine B capacity

3x + 1y ≤ 50

(16.67, 0)

(0, 50)

Finishing capacity

5x + 1y ≤ 70

(14, 0)

(0, 70)

Finding the corner points

The intersection of the Machine B constraint and the Finishing constraint forms one of the corners of the feasible region so the coordinates must be calculated.

Subtract the equation for Finishing from the equation for Machine B and solve for x

Machine B         3x + 1y = 50

Finishing         5x + 1y = 70[pic 1]

                -2x        = -20

                x = 10

Substitute the value for x into one of the constraints and solve for y

Machine B         3(10)+ 1y = 50

                30+ 1y = 50

                y = 20

        

Determine which corner point maximizes weekly profit[pic 2]

(0,0)        7(0) + 2(0) = $0/week                (10, 20)        7(10) + 2(20) = $110/week

(0,50)        7(0) + 2(50) = $100/week        (14, 0)        7(14) + 2(0) = $98/week


[pic 3][pic 4][pic 5]


A produce grower is purchasing fertilizer containing three nutrients: A, B and C.  The minimum weekly requirements are 80 units of A, 120 of B, and 240 of C.  There are two popular blends of fertilizer on the market.  Blend I costing $8 per bag, contains 2 units of A, 6 of B and 4 of C.  Blend II costing $10 per bag, contains 2 units of A, 2 of B and 12 of C.  How many bags of each blend should the grower buy each week to minimize the cost of meeting the nutrient requirements?

Decision Variables:

        x = Number of bags of Blend I the grower should purchase per week

y = Number of bags of Blend II the grower should purchase per week

Objective Function:

        Minimize cost        $8x + $10y

Subject to:

Minimum weekly requirement for Nutrient A:        2x + 2y ≥ 80

Minimum weekly requirement for Nutrient B        6x + 2y ≥ 120

Minimum weekly requirement for Nutrient C        4x + 12y ≥ 240

Non-negativity                                        x, y ≥ 0

Calculations:

Graphing the constraints:

The x- and y- intercepts for the of the constraints are

Constraint

x-intercept

y-intercept

Nutrient A

2x + 2y ≥ 80

(40, 0)

(0, 40)

Nutrient B

6x + 2y ≥ 120

(20, 0)

(0, 60)

Nutrient C

4x + 12y ≥ 240

(60, 0)

(0, 20)

Finding the corner points

The intersection of the Nutrient A and Nutrient B constraints forms one of the corners of the feasible region so the coordinates must be calculated.

Subtract the equation for Nutrient B from the equation for Nutrient A and solve for x

Nutrient A         2x + 2y = 80

Nutrient B         6x + 2y = 120

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