This problem is intended to guide you
through transposing a standard minimization problem
into a standard maximization problem.
We won't actually be solving.
The given minimization problem is called a primal problem,
and we'll be determining
the corresponding maximization problem,
which is called the dual problem.
Notice how we do have a standard minimization problem,
because all the variables are positive,
and all the constraints are in the form of
being greater than or equal to some non-negative value.
We'll be using these first two inequalities
and the objective function
to set up a matrix without using slack variables.
We'll use the coefficients as the entries into our matrix.
Using our first two inequalities,
we can determine the first two rows of our matrix,
and then we'll use the objective function
for the third row.
Looking at the first inequality,
notice how the coefficients would be
three, six, one, and four,
which is the first row of our matrix,
so we have three, six, one, four.
For the second row, we'd have
eight, one, four, and three,
so eight, one, four, and three.
Notice so far, the first column represents
the coefficients of x sub one.
Notice the first column represents
the coefficients of x sub one,
The second column represents the coefficients of x sub two,
and the third, the coefficients of x sub three.
Because of this, for the objective function,
we want to the coefficients of the x terms first,
and then the coefficient of c.
The third row is going to be 80, 25, 70,
and then one for the coefficient of c.
Again, we have 80,
25,
70 and one.
The next step, we want to transpose this matrix,
which means the rows will become the columns.
This first row will be the first column,
the second row will be the second column,
and the third row will be the third column.
The first column will be three, six, one, four.
The second column will be eight, one, four, three,
and the third column will be 80, 25, 70, and one.
This transposed matrix
is the matrix that we use
to set up the dual problem,
or the corresponding maximization problem.
Here's the same matrix,
and now we'll use this
to set up our max.
Here's the same matrix.
The first three rows will give us our constraints.
The last row will give us our objective function.
Notice how we're told to use the variables
y sub one, y sub two, and so on.
Looking at the first row,
this indicates that our first constraint will be
Three times y sub one
plus eight times y sub two
must be less than or equal to 80.
Notice how, because we have a maximization problem,
we're now using less than or equal to inequality symbols.
The second row would give us the constraint
six times y sub one plus one times y sub two,
or just y sub two
is less than our equal to 25.
The third row would give us the constraint
one times y sub one, or y sub one,
plus four times y sub two
must be less than or equal to 70.
Now, the last row gives our objective function,
which would be four times y sub one
plus three times y sub two
equals, let's say, p,
or p equals
four times y sub one
plus three times y sub two.
We should also include the constraints,
that y sub one and y sub two are non-negative,
so let's also say that y sub one
is greater than or equal to zero,
and y sub two is also greater than or equal to zero.
The next step would be to set up the initial tableau,
but we're going to stop here.
This is the dual problem.
I hope you found this helpful.