SOLUTION: Set up the simplex matrix used to solve this linear programming problem. all variables are nonnegative. Maximize {{{f=3x+5y+11z}}} subject to {{{system(2x+3y+4z<=60, x+4y+z<=48,

Introduce slack variables , , and ,
to "take up the slack" between the left sides
and the right sides of the first three and the
objective function, the one to maximize, at the bottom: 



Rearrange the terms in the bottom equation
and put in 0 and 1 coefficients of all the 
variables:

   

Then we set up the matrix with petitions as follows:


  
That matrix is called "the initial tableau".

That's all you were asked for, but I thought
I'd show you how to solve it.

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The numbers on the bottom left, 
all but the last one, are called "indicators",
Pick the indicator that is the most negative.  
That is, the . It is in column 3, so 
we call column 3 the "pivot column".

Next divide each of the three numbers above it into the number
at the far right of the row it's in. That is we do these
three divisions:

  15    48        48
4)60, 1)48, and 1)48 

The smallest of these is 15, obtained from row 1,
So we call row 1 the "pivot row".

And we call the number in both the pivot row and the
pivot column the "pivot element", which is 4.

Now we divide the pivot row by that number, 4, to make 
the pivot element become 1.



Now we use the pivot row to get 0's everywhere
else in the pivot column.

Then we get a 0 where the 1 is in row 2 by multiplying
-1 times the pivot row to row 2. 
 
Then we get a 0 where the 1 is in row 2 by multiplying
-1 times the pivot row to row 3.

Then we get a 0 where the -11 is in the bottom row by 
multiplying 11 times the pivot row to the bottom row.

Now we have:



Usually some of the indicators on the bottom
row will still be negative, and we would have 
to repeat the operation until there were no
negative indictors on the bottom row.  However 
this particular problem became solved with just 
one operation, so we convert this into a system 
of equations:






Now look at the bottom equation:



Solve for 



Since none of the variables can be negative,
f will be the largest value it can possibly be
when nothing is subtracted from the 165, so 
that maximum is reached when ,, and  are 
all taken to be . So the system now becomes



or



So  has a maximum value of  when
, , and 

Edwin