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WINSTONCH [101]
3 years ago
5

Donna de paul is raising money for the homeless. She discovers that each church group requires 2 hours of let her writing and 1

hour of follow-up while for each labor union she needs 2 hours of letter written and 3 hours of follow-up. Donna can raise $150 from each church group and $200 from each Union Local and she has a maximum of 20 hours of letters written and a maximum of 16 hours of follow-up available per month. Determine the most profitable mixture of group she should contact in the most money she can lose in a month.
z=()x1+()x2
Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
5 0

Answer:

x  =  7

y  =  3

z (max)  =  4950/3  =  1650

Step-by-step explanation:

Let call  

x numbers of church goup and

y numbers of Union Local

Then  

First contraint

2*x + 2*y ≤ 20

Second one

1*x + 3*y ≤ 16

Objective Function

z = 150*x + 200*y

Then the system is

z = 150*x + 200*y To maximize

Subject to:

2*x + 2*y ≤ 20

1*x + 3*y ≤ 16

x ≥ 0 y ≥ 0

We will solve by using the Simplex method

z - 150 *x - 200*y = 0

2*x + 2*y + s₁ = 20  

1*x + 3*y + 0s₁ + s₂ = 16

First Table

z           x          y          s₁           s₂          Cte

1        -150    -200       0            0     =      0

0          2         2           1            0     =    20

0          1          3          0            1      =     16

First iteration:

Column pivot   ( y column ) row pivot (third row) pivot 3

Second table

z           x                y          s₁           s₂                Cte

1       -250/3           0          0         200/3    =    3200/3

0      - 4/3               0          -1           2/3       =    -20/3

0         1/3               1            0           1/3       =    -20/3

Second iteration:

Column pivot  ( x column ) row pivot  (second row)  pivot  -4/3

Third table

z           x                y          s₁           s₂                Cte

1            0               0      750/12    700/6    =   4950/3

0            1                0        3/4          -1/2     =    7

0            0                1         -1/4          1/2     =  9/3

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Select all the points that are on the line through (0 5) and (2 8) A.(5 11) B.(5 10) C.(6 14) D.(30 50) E.(40 60)
mihalych1998 [28]

Answer:

Options C. and D. are correct.

Step-by-step explanation:

Let (x_1,y_1)=(0,5),\,(x_2,y_2)=(2,8)

Equation of a line is given by y-y_1=(\frac{y_2-y_1}{x_2-x_1})(x-x_1)

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Put  (x,y)=(5,11)

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Put  (x,y)=(5,10)

3(5)-2(10)+10=15-20+10=5\neq 0

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3(6)-2(14)+10=18-28+10=0

Put (x,y)=(30,50)

3(30)-2(50)+10=90-100+10=0

Put (x,y)=(40,60)

3(40)-2(60)+10=0\\120-120+10=10\neq 0

So, points (6,14),\,(30,50) satisfy the equation 3x-2y+10=0

Therefore,

points (6,14),\,(30,50) lie on the line through (0,5) and (2,8)

Options C. and D. are correct.

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Degger [83]

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iogann1982 [59]

Answer:

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6 0
3 years ago
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Zinaida [17]
I think the answer is c sorry if I’m wrong
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