Section 54
(14)
Maximize

P = 15x_{1} + 20x_{2} 
subject to ...

2x_{1} + x_{2} ≤ 9 
x_{1} + x_{2} ≤ 6  
x_{1} + 2x_{2} ≤ 10  
x_{1}, x_{2}, ≥ 0 
There are three problem constraints, so there will be three slack variables. The system looks this way:
2x_{1}

+ x_{2}

+ s_{1}

= 9  
x_{1}

+ x_{2}

+ s_{2}

= 6  
x_{1}

+ 2x_{2}

+ s_{3}

= 10  
15x_{1}

 20x_{2}

+ P

= 0 
Now this system is put into a simplex tableau.
x_{1}

x_{2}

s_{1}

s_{2}

s_{3}

P


s_{1}

2

1

1

0

0

0



9

9/1 = 9  
s_{2}

1

1

0

1

0

0



6

6/1 = 6  
s_{3}

1

2

0

0

1

0



10

10/2 = 5 (smallest)  









15

20

0

0

0

1



0

From this tableau we can read off our starting BFS of x_{1}=0,
x_{2}=0,
s_{1}=9, s_{2}=6, s_{3}=10, P=0.
There are negative indicatiors so we pick the pivot column (blue) and pivot
row (salmon).
Now we must do row operations to turn the pivot entry into a 1 and get 0's elsewhere
in the pivot column.
1/2R_{3} > R_{3} 

R_{3} + R_{1} > R_{1} R_{3} + R_{2} > R_{2} 20R_{3} + R_{4} > R_{4} 

There's still a negative indicator, so we choose a new pivot row and pivot
column, illustrated above. Again we must do the row operations to change the
pivot element to a 1 and the remaining elements in the pivot column to 0's.
Here comes the 1 that we want:
2R_{2} > R_{2} 

...and here come the zeros:
3/2R_{2} + R_{1} > R_{1} 2R_{2} + R_{3} > R_{3} 5R_{2} +_{ R4 > R4} 

Now there are no more negative indicators, a real cause for celebration! Reading off the final BFS from the tableau, we see that x_{1}=2, x_{2}=4, s_{1}=1, s_{2}=0, s_{3}=0, P=110.
Thus P is maximized at 110 when x_{1}=2 and x_{2}=4,
(20)
Maximize... 
P = 4x_{1}  3x_{2} + 2x_{3}

Subject to... 
x_{1} + 2x_{2}  x_{3} ≤ 5

3x_{1} + 2x_{2} + 2x_{3}
≤
22


x_{1}, x_{2, x3 }≥ 0

There are two problem constraints, so there will be two slack variables. The system looks this way:
x_{1}

+ 2x_{2}

 x_{3}

+ s_{1}

= 5


3x_{1}

+ 2x_{2}

+ 2x_{3}

+ s_{2}

= 22


4x_{1}

+ 3x_{2}

 2x_{3}

+ P

= 0

Now this system is put into a simplex tableau. The pivot column is blue and
the pivot row is salmon.
x_{1}

x_{2}

x_{3}

s_{1}

s_{2}

P


s_{1}

1

2

1

1

0

0



5

5/1 = 5  
s_{2}

3

2

2

0

1

0



22

22/3 ~7.3  









P

4

3

2

0

0

1



0

Luckily, the pivot entry is already 1, so we don't have to multiply the pivot row. We just need to get the remaining entries of the pivot column to be zeros.
3R_{1} + R_{2} > R_{2} 4R_{1} + R_{3} > R_{3} 

At this point x_{1} has become basic and s_{1} has become nonbasic (i.e. zero). You can read off the BFS x_{1} = 5 , x_{2} = 0, x_{3} = 0, s_{1} = 0, s_{2} = 7, P = 20.
But there's still a negative indicator (6) so we have to do another pivot operation. The pivot column is blue again. For the pivot row we have no choice other than row 2, because it contains the only nonnegative entry above the line in the pivot column. The pivot row is shown in salmon.
To start the second poivt operation, we must turn the pivot entry 5 into a 1:
1/5 R_{2} > R_{2} 

Next we need to get zeros for the remaining entries in the pivot column:
R_{2} + R_{1} > R_{1} 6 R_{2} + R_{3} > R_{3} 

There. There are no more negative indicators. We read off the following BFS:
x_{1} = 6.4, x_{2} = 0, x_{3}
= 1.4, s_{1} = 0, s_{2} = 0, P = 28.4
Solution: Maximize P at 28.4 by setting x_{1} = 6.4, x_{2} = 0, x_{3} = 1.4.
(39)
Let x_{1} = number of daytime ads, x_{2}
= number of primetime ads, and x_{3} = number of late night ads.
Then the total number of potential customers reached is 14000 x_{1}
+ 24000 x_{2} + 1800 x_{3}.
The total amount the chain is spending on the ads is 1000 x_{1}
+ 2000 x_{2} + 1500 x_{3} dollars.
The TV station says that the total number x_{1}
+ x_{2} + x_{3} of ads must not be greater than 15.
Thus we want to maximize the number of potential customers  P = 14000 x_{1} + 24000 x_{2} + 1800 x_{3} 
Subject to ...  1000 x_{1} + 2000 x_{2} + 1500 x_{3} ≤ 20000 
x_{1} + x_{2} + x_{3} ≤ 15  
x_{1}, x_{2}, x_{3} ≥ 0 
There are 2 problem contraints, so there will be 2 slack variables. The system becomes:
1000 x_{1}

+ 2000 x_{2 }

+ 1500 x_{3}

+ s_{1}

= 20000  
x_{1}

+ x_{2}

+ x_{3}

+ s_{2}

= 15  
14000 x_{1}

24000 x_{2}

 18000 x_{3}

+ P

= 0 
Now we set up the simplex tableau. The pivot column is the second column. The pivot row is the first row.

1/2000 R_{1} > R_{1} 

R_{1} + R_{,2} > R_{,2} 24000 R_{1} + R_{3} > R_{3} 

2R_{2} > _{2} 

1/2 R_{2} + R_{1} > R_{1} 2000 R_{2} + R_{3} > R_{3} 

Now there are no more negative indicators, so we are done. The potential customer contacts are maximized at P = 260000, when there are 10 daytime ads, 5 primetime ads, and 0 late night ads.
(41) First let's tabulate the data:
colonial

split level

ranch

available


acres 
0.5

0.5

1

30

capital ($) 
60,000

60,000

80,000

3,200,000

hours 
4,000

3,000

4,000

18,000

profit ($) 
20,000

18,000

24,000

We need to find out how many colonial, splitlevel, and ranch houses must be constructed to maximize profit. Thus,
Let x_{1} be number of colonial houses.
Let x_{2} be number of splitlevel houses.
Let x_{3} be number of ranch houses.
Then the profit which we want to maximize is P = 20,000x_{1} + 18,000x_{2}
+ 24,000x_{3}
Thus we want to maximize

P = 20,000x_{1} + 18,000x_{2} + 24,000x_{3} 
subject to ...

0.5x_{1} + 0.5x_{2} + x_{3} ≤ 30 
60,000x_{1} + 60,000x_{2} + 80,000x_{3 }≤ 3,200,000  
4,000x_{1} + 3,000x_{2} + 4,000x_{3} ≤ 180,000  
x_{1}, x_{2}, x_{3} ≥ 0 
Those are some mighty big numbers. There's no harm in dividing some of the lines by 1000, so let's do that to get the revised problem
maximize

P = 20x_{1} + 18x_{2} + 24x_{3} 
subject to ...

0.5x_{1} + 0.5x_{2} + x_{3} ≤ 30 
60x_{1} + 60x_{2} + 80x_{3 }≤ 3,200  
4x_{1} + 3x_{2} + 4x_{3} ≤180  
x_{1}, x_{2}, x_{3} ≥ 0 
That looks better. Remember though, P now represents profit in thousands of dollars.
There are three problem contraints, so there will be 3 slack variables. The system becomes:
0.5x_{1}

+ 0.5x_{2 }

+ x_{3}

+ s_{1}

= 30  
60x_{1}

+ 60x_{2}

+ 80x_{3}

+ s_{2}

= 3,200  
4x_{1}

+ 3x_{2}

+ 4x_{3}

+ s_{3}  = 180  
20x_{1}

18x_{2}

 24x_{3}

+ P

= 0 
Now we set up the simplex tableau.
x_{1}

x_{2}

x_{3}

s_{1}

s_{2}

s_{3}

P


s_{1}

0.5

0.5

1

1

0

0

0



30

30/1 = 30 (smallest)  
s_{2}

60

60

80

0

1

0

0



3,200

3200/80=40  
s_{3}

4

3

4

0

0

1

0



180

180/4 = 45  










P

20

18

24

0

0

0

1



0

The pivot element is already 1. That's good. We just need to get zeros under
the pivot entry.
80R_{1} + R_{2} > R_{2}
4R_{1} + R_{3} > R_{3}
24R_{1} + R_{4} > R_{4}
x_{1}

x_{2}

x_{3}

s_{1}

s_{2}

s_{3}

P


x_{3}

0.5

0.5

1

1

0

0

0



30

30/0.5 = 60  
s_{2}

20

20

0

80

1

0

0



800

800/20 = 40  
s_{3}

2

1

0

4

0

1

0



60

60/2 = 30 (smallest)  










P

8

6

0

24

0

0

1



720

1/2R_{3} > R_{3}
0.5

0.5

1

1

0

0

0



30


20

20

0

80

1

0

0



800


1

0.5

0

2

0

0.5

0



30












8

6

0

24

0

0

1



720

1/2R_{3} + R_{1} > R_{1}
2R_{3} + R_{2} > R_{2}
8R_{3} + R_{4} > R_{4}
x_{1}

x_{2}

x_{3}

s_{1}

s_{2}

s_{3}

P


x_{3}

0

0.25

1

2

0

.25

0



15

15/.25=60  
s_{2}

0

1

0

40

1

10

0



200

200/10=20(smallest)  
x_{1}

1

0.5

0

2

0

0.5

0



30

30/0.5=60  










P

0

2

0

8

0

4

1



960

There's still a negative indicator above, so it's time for another pivot operation.
The pivot row and column are indicated above.
1/4R_{2} + R_{1} > R_{1}
1/2R_{2} + R_{3} > R_{3}
2R_{2} + R_{4} > R_{4}
x_{1}

x_{2}

x_{3}

s_{1}

s_{2}

s_{3}

P


x_{3}

0

0

1

3

0.25

0

0



10


x_{2}

0

1

0

4

1

1

0



20


x_{1}

1

0

0

0

0.5

1

0



20












P

0

0

0

0

2

2

1



1000

There are no longer any negative indicators, so we can now read off the final solution. There will be a profit of a cool million by building 10 ranch houses, 20 splitlevels, and 20 colonial styles. (recall that the P=1000 means the profit is 1000 thousand dollars, i.e. one million dollars.)
Frankly, though, this kind of development is bad for the environment. Instead of building all those houses, the developer should donate his capital to the Sierra Club.