I recommend checking out this page to help refresh your memory how the Simplex Method works.
https://people.richland.edu/james/ictcm/2006/simplex.html
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We wish to maximize P = 10x+40y
The constraints are:
Focus on this subset of constraints
Convert to their corresponding equation format
where s,t,u are the slack variables mentioned. They are nonnegative real numbers.
Let's have the slack variables present in each equation.
A coefficient of '1' means the slack variable is actually present in that particular equation.
A coefficient of '0' means it is not present.
This system
is equivalent to
Then an equation like
2x+1y+1s+0t+0u = 100
will have the coefficients: 2,1,1,0,0
which will make up part of row1 in the simplex tableau.
This is our starting simplex tableau
where C represents the right-hand-side constants of each equation mentioned above.
The bottom row represents the equation -10x-40y+P = 0
This is a rephrasing of P = 10x+40y where everything is to one side.
Write 0's in the slack entries of the bottom row
This is because
-10x-40y+P = 0
is the same as
-10x-40y+0s+0t+0u+1P = 0
Currently this tableau is not optimized because we have negative items in the bottom row.
Select the negative item furthest from zero, which is -40. Highlight or circle this entire column. This is the current pivot column.
To find the pivot row, we'll need to divide each C value over each item in that pivot column.
row1: 100/1 = 100
row2: 500/10 = 50
row3: 2000/10 = 200
The smallest ratio will point us to the pivot row. In this case, that would be row 2.
The item "10" in this pivot column and pivot row must be made into a "1".
We do this by multiplying everything in this row by 1/10, i.e. divide everything in this row by 10.
I'll write the notation (1/10)*R2 --> R2 to indicate "multiply everything in row 2 (aka R2) by 1/10, then store the result in R2".
This is what we get after that row operation
Now we must zero out everything but this '1' in that column.
This pivot item is highlighted in yellow.
We are clearing out this column so to speak.
Use row operations to get this done.
Something like R1-R2 --> R1 means we subtract the corresponding entries of row1 and row2 in that order, then store the result in row1.
For more information, check out the concept of "row reduction".
This is the result of clearing out the pivot column mentioned
The row operations performed are shown on the right.
At this point, there aren't any negative items in the bottom row. Therefore, this tableau is optimized.
The variables y, s, u, and P are considered basic variables.
This is because they contain zeros in the non-pivot entries for the given column. These columns have been "cleared out" so to speak.
Refer to the link I posted above.
The nonbasic variables are x and t, as they don't have such zeros.
Nonbasic variables will take on the value of zero.
Therefore: x = 0 and t = 0
Going back to the basic variables, we have
y = 50
s = 50
u = 1500
P = 2000
Each value mentioned is in the column C at the far right side.
We have u = 1500 for instance because of the '1' in that corresponding row.
To put everything together:
x = 0, y = 50
s = 50, t = 0, u = 1500
P = 2000
The nonzero slack values for variable s and variable u indicate that the point (x,y) = (0,50) is not on the boundaries of 2x+y≤100 and 30x+10y≤2000 respectively.
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Answer:
The max is P = 2000
It occurs when x = 0 and y = 50
You can use graphing to confirm the answer.