Section 5-4

(14)

There are three problem constraints, so there will be three slack variables. The system looks this way:

Now this system is put into a simplex tableau.

From this tableau we can read off our starting BFS of x₁=0, x₂=0, s₁=9, s₂=6, s₃=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.

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:

...and here come the zeros:

Now there are no more negative indicators, a real cause for celebration! Reading off the final BFS from the tableau, we see that x₁=2, x₂=4, s₁=1, s₂=0, s₃=0, P=110.

Thus P is maximized at 110 when x₁=2 and x₂=4,

(20)

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

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

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.

At this point x₁ has become basic and s₁ has become nonbasic (i.e. zero). You can read off the BFS x₁ = 5 , x₂ = 0, x₃ = 0, s₁ = 0, s₂ = 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:

Next we need to get zeros for the remaining entries in the pivot column:

There. There are no more negative indicators. We read off the following BFS:
x₁ = 6.4, x₂ = 0, x₃ = 1.4, s₁ = 0, s₂ = 0, P = 28.4

Solution: Maximize P at 28.4 by setting x₁ = 6.4, x₂ = 0, x₃ = 1.4.

(39)
Let x₁ = number of daytime ads, x₂ = number of primetime ads, and x₃ = number of late night ads.
Then the total number of potential customers reached is 14000 x₁ + 24000 x₂ + 1800 x₃.
The total amount the chain is spending on the ads is 1000 x₁ + 2000 x₂ + 1500 x₃ dollars.
The TV station says that the total number x₁ + x₂ + x₃ of ads must not be greater than 15.

There are 2 problem contraints, so there will be 2 slack variables. The system becomes:

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

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:

We need to find out how many colonial, split-level, and ranch houses must be constructed to maximize profit. Thus,

Let x₁ be number of colonial houses.
Let x₂ be number of split-level houses.
Let x₃ be number of ranch houses.
Then the profit which we want to maximize is P = 20,000x₁ + 18,000x₂ + 24,000x₃

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

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:

Now we set up the simplex tableau.

The pivot element is already 1. That's good. We just need to get zeros under the pivot entry.

-80R₁ + R₂ --> R₂
-4R₁ + R₃ --> R₃
24R₁ + R₄ --> R₄

1/2R₃ --> R₃

-1/2R₃ + R₁ --> R₁
-2R₃ + R₂ --> R₂
8R₃ + R₄ --> R₄

There's still a negative indicator above, so it's time for another pivot operation. The pivot row and column are indicated above.
-1/4R₂ + R₁ --> R₁
-1/2R₂ + R₃ --> R₃
2R₂ + R₄ --> R₄

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 split-levels, 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.