A Farmer has 600 acres that he can plant in wheat or barley. He also has 6000 pounds of fertilizer.
Wheat takes 15 pounds of fertilizer per acre and barley takes 12 pounds of fertilizer per acre. The
farmer makes a profit of $100 for each acre of wheat and $120 for each acre of barley. Maximize the
farmer’s profit.
The unknowns are the acres of wheat and acres of barley. Let x1 be the number of acres of wheat and
x2 be the number of acres of barley. (This choice of variables is arbitrary and you could choose
differently, and that is OK).
The limitations on x1 and x2 are expressed as follows.
x1 is not negative: x1 >=0. (i.e. the farmer cannot plant negative acres of wheat)
x2 is not negative: x2 >=0. (i.e. the farmer cannot plant negative acres of barley)
x1+x2 <= 600 (i.e. the farmer cannot plant more than 600 acres altogether)
Each acre of wheat uses 15 pounds of fertilizer. So the fertilizer used for wheat is 15x1.
Each acre of barley uses 12 pounds of fertilizer. So the fertilizer used for barley is 12x2.
The total fertilizer used is at most 6000 pounds so 15x1 + 12x2 <= 6000.
The summary constraints are:
x1 >=0
x2 >=0
x1 + x2 <= 600
15x1 + 12x2 <= 6000
The profit is P = 100x1+ 120x2
For the simplex algorithm we rewrite these constraints as follows:
x1 + x2 + s1 = 600
15x1 + 12x2 +s2 = 6000
-100x1 + -120x2 + P = 0
Giving this simplex table:
X1
X2
S1
S2
P
Constant
1
1
1
0
0
600
15
12
0
1
0
6000
-100
-120 0
0
1
0
We can interpret this table as a feasible solution. Any variable which has numbers other than 0 and one
1 in its column has a value 0. So the interpretation of the initial table is
{x1=0; x2=0;s1=600;s2=6000;P=0}. (No acres are planted, giving a profit of $0.
We find the most negative number in the bottom row: (-120) and select that column
X1
1
15
-100
X2
1
12
-120
S1
1
0
0
S2
0
1
0
P
0
0
1
Constant
600
6000
0
We do this, because this represents the change which has the most impact on the profit.
Next we find the ratios between that column and the constant column.
600/1 = 600
6000/12 = 500
(ignore 0 or negative ratios)
Pick the smallest ratio = 500 This selects the pivot element. We pick the smallest ratio, because if we
selected another ratio we would violate one of the other constraints after we pivoted.
X1
1
15
-100
X2
1
12
-120
S1
1
0
0
S2
0
1
0
P
0
0
1
Constant
600
6000
0
Divide the row of the pivot element by the pivot element to get a pivot of 1.
X1
1
15/12
-100
X2
1
1
-120
S1
1
0
0
S2
0
1/12
0
P
0
0
1
Constant
600
6000/12=500
0
Eliminate all other numbers in that column:
R1 – R2 -> R1
And
R3 + 120 R2 -> R3
X1
1-15/12=-3/12
15/12
-100+120*15/12=50
X2
1-1=0
1
-120+120=0
S1
1-0 =1
0
0+0=0
S2
0-1/12 =-1/12
1/12
0+120*12=10
P
0-0=0
0
1+0=1
Constant
600-500=100
6000/12=500
0+120*500=6000
X1
-3/12
15/12
50
X2
0
1
0
S1
1
0
0
S2
-1/12
1/12
10
P
0
0
1
Constant
100
500
6000
Interpreting the result: x1 has numbers in the column other than a single 1 and the rest 0’s so
x1=0;
x2= 500;
s1=100;
s2=0;
P=6000
There are no more negative numbers in the bottom row, so we are finished.
The maximum profit for the farmer occurs if he plants 500 acres of barley, leaves 100 acres fallow, and
uses all the fertilizer. The profit is $6000
x1 = acres of wheat = 0;
x2 = acres of barley = 500;
s1 = unused land = 100 acres;
s2 = unused fertilizer = 0;
P = profit = $6000
On the TI 83/84:
A. Enter the simplex table (tableau) into the calculator as a matrix:
B. (2nd)(Matrix)->Edit->3X6->
1
1
15
12
-100 -120
(2nd)(QUIT)
1
0
0
0
1
0
0
0
1
600
6000
0
C. Pick the pivot element (12) as above.
D. Divide the pivot row by the pivot element:
(2nd)(Matrix) (MATH)*row(1/12,[A],1)
(2nd)(QUIT)
E. Subtract the new second row from the first row;
(2nd)(Matrix) (MATH)*row+(-1, (2nd)ANS,2,1)
(2nd)(QUIT)
F. Add 120 times the second row to the third row
(2nd)(Matrix) (MATH)*row+(120, (2nd)ANS,2,3)
(2nd)(QUIT)
Examine the resulting third row. Since it has no negative numbers we are finished. We will
store our answer (STO) (2nd) (Matrix) [B].
Note that we could have stored each step along the way instead of basing the next calculation
on the previous result.
This would be useful if we wanted to use the calculator for computing the ratios.