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LenKa [72]
3 years ago
8

Better Products, Inc., manufactures three products on two machines. In a typical week, 40 hours are available on each machine. T

he profit contribution and production time in hours per unit are as follows:
Category Product 1 Product 2 Product 3
Profit/unit $30 $50 $20
Machine 1 time/unit 0.5 2.0 0.75
Machine 2 time/unit 1.0 1.0 0.5
Two operators are required for machine 1; thus, 2 hours of labor must be scheduled for each hour of machine 1 time. Only one operator is required for machine 2. A maximum of 100 labor-hours is available for assignment to the machines during the coming week. Other production requirements are that product 1 cannot account for more than 50% of the units produced and that product 3 must account for at least 20% of the units produced.
How many units of each product should be produced to maximize the total profit contribution?
Product # of units
1
2
3
What is the projected weekly profit associated with your solution?
Profit = $
How many hours of production time will be scheduled on each machine? If required, round your answers to two decimal places.
Machine Hours Schedule:
Machine 1 Hours
Machine 2 Hours
What is the value of an additional hour of labor? If required, round your answers to two decimal places.
$
Assume that labor capacity can be increased to 120 hours. Develop the optimal product mix, assuming that the extra hours are made available.
Product # of units
1
2
3
Profit = $
Would you be interested in using the additional 20 hours available for this resource?
Mathematics
1 answer:
Kaylis [27]3 years ago
6 0

Answer:

z (max)  =  1250 $

x₁  = 25    x₂  =  0   x₃  =  25

Step-by-step explanation:

                                Profit $    mach. 1      mach. 2

Product 1     ( x₁ )       30             0.5              1

Product 2    ( x₂ )       50             2                  1

Product 3    ( x₃ )       20             0.75             0.5

Machinne 1 require  2 operators

Machine   2 require  1  operator

Amaximum of  100 hours of labor available

Then Objective Function:

z  =  30*x₁  +  50*x₂  +  20*x₃      to maximize

Constraints:

1.-Machine 1 hours available  40

In machine 1    L-H  we will need

0.5*x₁  +  2*x₂  + 0.75*x₃  ≤  40

2.-Machine 2   hours available  40

1*x₁  +  1*x₂   + 0.5*x₃   ≤  40

3.-Labor-hours available   100

Machine 1     2*( 0.5*x₁ +  2*x₂  +  0.75*x₃ )

Machine  2       x₁   +   x₂   +  0.5*x₃  

Total labor-hours   :  

2*x₁  +  5*x₂  +  2*x₃  ≤  100

4.- Production requirement:

x₁  ≤  0.5 *( x₁ +  x₂  +  x₃ )     or   0.5*x₁  -  0.5*x₂  -  0.5*x₃  ≤ 0

5.-Production requirement:

x₃  ≥  0,2 * ( x₁  +  x₂   +  x₃ )  or    -0.2*x₁  - 0.2*x₂ + 0.8*x₃   ≥  0

General constraints:

x₁  ≥   0       x₂    ≥   0       x₃     ≥   0           all integers

The model is:

z  =  30*x₁  +  50*x₂  +  20*x₃      to maximize

Subject to:

0.5*x₁  +  2*x₂  + 0.75*x₃  ≤  40

1*x₁  +  1*x₂   + 0.5*x₃       ≤  40

2*x₁  +  5*x₂  +  2*x₃        ≤  100

0.5*x₁  -  0.5*x₂  -  0.5*x₃  ≤ 0

-0.2*x₁  - 0.2*x₂ + 0.8*x₃   ≥  0

x₁  ≥   0       x₂    ≥   0       x₃     ≥   0           all integers

After 6 iterations with the help of the on-line solver AtomZmaths we find

z (max)  =  1250 $

x₁  = 25    x₂  =  0   x₃  =  25

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B.) 2083

C.) 3972

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A = A0e^rt

A0 = initial population

A = final population

r = growth rate ; t = time

1)

Using the year 1750 and 1800

Time, t = 1800 - 1750 = 50 years

Initial population = 790

Final population = 980

Let's obtain the growth rate :

980 = 790e^50r

980/790 = e^50r

Take the In of both sides

In(980/790) = 50r

0.2155196 = 50r

r = 0.2155196/50

r = 0.0043103

Using this rate, let predict the population in 1900

t = 1900 - 1750 = 150 years

A = 790e^150*0.0043103

A = 790e^0.6465588

A = 1508.0788 ; 1508 million people

In 1950;

t = 1950 - 1750 = 200

A = 790e^200*0.0043103

A = 790e^0.86206

A = 1870.7467 ; 1870 million people

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Exponential model. For 1800 and 1850

Initial, 1800 = 980

Final, 1850 = 1260

t = 1850 - 1800 = 50

Using the exponential format ; we can obtain the rate :

1260 = 980e^50r

1260/980 = e^50r

Take the In of both sides

In(1260/980) = 50r

0.2513144 = 50r

r = 0.2513144/50

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Using the model ; The predicted population in 1950;

In 1950;

t = 1950 - 1800 = 150

A = 980e^150*0.0050262

A = 980e^0.7539432

A = 2082.8571 ; 2083 million people

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r = 0.4392319/50

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Using the model ; The predicted population in 2000;

In 2000;

t = 2000 - 1900 = 100

A = 1650e^100*0.0087846

A = 1650e^0.8784639

A = 3971.8787 ; 3972 million people

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