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Usimov [2.4K]
2 years ago
7

. A 1000-gallon tank initially contains a mixture of 450 gallons of cola and 50 gallons of cherry syrup. Cola is added at the ra

te of 8 gallons per minute, and cherry syrup is added at the rate of 2 gallons per minute. At the same time, a well mixed solution of cherry cola is withdrawn at the rate of 5 gallons per minute. What percentage of the mixture is cherry syrup when the tank is full
Mathematics
1 answer:
marishachu [46]2 years ago
6 0

The tank is filled up at a total rate of 10 gal/min, and currently contains 500 gal of solution. It can hold up to 1000 gal, so it will take time <em>t</em> to fill it up completely:

500 gal = (10 gal/min) <em>t</em>

<em>t</em> = (500 gal) / (10 gal/min)

<em>t</em> = 50 min

In that time,

(8 gal/min) (50 min) = 400 gal of cola

and

(2 gal/min) (50 min) = 100 gal of cherry syrup

are added to the tank, so that when it's full, the tank contains 850 gal of cola and 150 gal of syrup.

So 150 gal of the total 1000 gal of solution is cherry syrup, or (150 gal)/(1000 gal) = 15/100 = 15% of the mixture.

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z (max)  =  1250 $

x₁  = 25    x₂  =  0   x₃  =  25

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                                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

6 0
3 years ago
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