Here the goal is to find values for x (number of Soy Joy Smoothies) and y (number of Vitamin Boost Smoothies) that will yield the highest value of R in the equation
(a) R = 2.75x + 3.25y.
One is constrained by the restrictions:
(b) 2x + 1y ≤ 100
(c) 1x + 4y ≤ 120
Both x and y are greater than or equal to 0.
Write the Initial Tableau shown below. Note the placement of the coefficients from Expressions (a), (b), & (c) in the Initial Tableau; 2.75 and 3.25 must have negative signs in the tableau to achieve correct results.
Initial Tableau------x------------y-----------s-----------t----------R-----------b
s-----------------------2-----------1-----------1-----------0----------0---------100----ROW1
t-----------------------1-----------4------------0-----------1----------0---------120---ROW2
R-----------------(-2.75)-----(-3.25)--------0-----------0----------1-----------0-----ROW3
The smallest negative number in ROW3 is -3.25 and it stands below 1 & 4 in the y column. Trace 1 and 4 to their corresponding entries in the b column which are 100 & 120. Divide 100 by 1 and 120 by 4 to obtain 100 and 30. 30 is the smaller of these quotients, so trace upward from -3.25 and left from 120 to meet at 4. This makes 4 the Pivot Entry. Also, let t in the far left column "depart" and replace it with y, the "entering variable".
Next divide all entries on ROW2 by 4 to obtain a 1 in this first Pivot Entry position. This gives the Second Tableau shown below:
Second Tableau------x------------y-----------s-----------t-------------R-----------b
s--------------------------2------------1-----------1------------0-----------0----------100-----ROW1
y--------------------------1/4----------1-----------0------------1/4---------0-----------30-----ROW2/4
R--------------------(-2.75)-----(-3.25)---------0------------0-----------1-------------0-----ROW3
Now add the negative of each entry in ROW2/4 to its corresponding entry in ROW1 and place each result into the "old" ROW1 slot. Also add the product of 3.25 and each entry in Row2/4 to its corresponding
entry in ROW3 and place each result into the "old" ROW3 slot. The aim here has been to come up with a y column of all zeroes except for the 1 in the Pivot Entry position. This gives the Third Tableau shown below:
Third Tableau---------x-------------y------------s------------t-----------R---------b
s-------------------------7/4-----------0------------1---------(-1/4)--------0--------70----(-ROW2/4)+ROW1---------[ROW-A]
y-------------------------1/4-----------1-------------0-----------1/4---------0--------30----ROW2/4----------------------[ROW-B]
R--------------------(-1.9375)------ 0-------------0--------0.8125-------1------97.5--(3.25×ROW2/4)+ROW3----[ROW-C]
There is still a negative entry in the bottom row of the Third Tableau, so next divide 70 by 7/4 in the top row and divide 30 by 1/4 (x column) in the middle row to yield 40 and 120. Since 40 is the smaller quotient, trace upward from -1.9375 and left from 70 to meet at 7/4 in the x column. This is the new Pivot Entry. Let s in the far left column "depart" and replace it with x, the "entering variable". Rename the rows as [ROW-A], [ROW-B], and [ROW-C] for convenience. Then multiply [ROW-A] by 4/7 to obtain the top row of the Fourth Tableau shown below:
Fourth Tableau---------x-----------y------------s------------t-----------R---------b
x---------------------------1-----------0------------4/7---------(-1/7)------0--------40----4[ROW-A]/7
y-------------------------1/4-----------1-------------0-----------1/4---------0--------30-----[ROW-B]
R--------------------(-1.9375)------ 0-------------0--------0.8125-------1------97.5----[ROW-C]
Develop again a column (this time the x column) that is all zeroes except for the 1 in the new Pivot Entry position. Write the Fifth Tableau by adding -0.25×4[ROW-A]/7 to [ROW-B] and placing the results as the new middle row. Add 1.9375×4[ROW-A]/7 to [ROW-C] and place the results as the new bottom row:
Fifth Tableau------------x-----------y------------s----------------t-----------R---------b
x---------------------------1-----------0------------4/7----------(-1/7)---------0--------40-----4[ROW-A]/7
y---------------------------0-----------1----------(-1/7)------------2/7---------0--------20-----(-0.25)×4[ROW-A]/7+[ROW-B]
R--------------------------0-----------0----------31/28---------15/28--------1-------175-----1.9375×4[ROW-A]/7+[ROW-C]
Now there are no negative values left in the bottom row of this final Tableau. Trace from x in the far left column to 40 in the b column. Trace from y in the far left column to 20 in the b column. Finally, trace from
R in the far left column to 175 in the b column.
Revenue is then maximized to $175.00 when 40 Soy Joy Smoothies and 20 Vitamin Boost Smoothies are sold.
One can also graph 2x + y = 100 and x + 4y = 120 and obtain an irregular quadrilateral formed by these two lines and the coordinate axes. The quadrilateral will have corner points at (x,y) equal to (0,0), (0,30), (40,20) and (50,0). Evaluation of R = 2.75x + 3.25y at these corner points will again give the highest value of R as 175 at (x,y) = (40,20).
