There are different methods to solve a linear programming problem, but one common approach is the simplex method. Applying this method to the given problem, we can write it in standard form:

minimize z = 8x1 - 2x2

subject to:

-4x1 + 2x2 + s1 = 1
5x1 - 4x2 + s2 = 3
x1, x2, s1, s2 ≥ 0

where s1 and s2 are slack variables that represent the unused resources in each constraint.

We can then use the simplex method to iterate through the following steps:

1. Choose an initial feasible solution by setting the slack variables to their respective right-hand side values and the non-basic variables to zero. In this case, we have:

x1 = 0, x2 = 0, s1 = 1, s2 = 3

2. Compute the reduced costs of the non-basic variables. These are the coefficients of the objective function minus the sum of the products of each coefficient with its corresponding column in the constraint matrix. In this case, we have:

rc1 = 8 - (-4)*0 - 5*0 = 8
rc2 = -2 - 2*0 + 4*0 = -2

3. If all reduced costs are non-negative, the current solution is optimal. Otherwise, choose a non-basic variable with negative reduced cost as the entering variable. In this case, x2 has negative reduced cost, so it will enter the basis.

4. Compute the ratios of the right-hand side values to the coefficients of the entering variable in each constraint. If a ratio is negative or zero, the corresponding constraint is said to be degenerate and the method may fail or cycle. In this case, both ratios are positive:

r1 = 1/2 = 0.5
r2 = 3/(-4) = -0.75

5. Choose the constraint with the smallest positive ratio as the leaving constraint. In this case, the entering variable x2 corresponds to the first constraint, so we choose that one.

6. Perform a pivot operation to make the entering variable basic and the leaving variable non-basic. To do this, divide the leaving constraint by the coefficient of the entering variable in that constraint (2), and subtract from it a multiple of the entering constraint that eliminates the entering variable from that constraint. The result is a new set of constraints and a new objective function:

x2 = 1/2x1 - 1/2s1 + 1/2
x2 = 5/4 - 1/4x1 - 3/4s2
z = 4x1 + 2s1 + 16/3s2 - 5/2

7. Go back to step 2 and repeat until an optimal solution is reached.

After a few more iterations, we find that the optimal solution is:

x1 = 1/2, x2 = 1/2, s1 = 0, s2 = 0

with objective value z = 1.5.

Therefore, the best feasible solution for the given LP problem is x1 = 1/2, x2 = 1/2, with minimum value of z = 1.5. This means that we can produce the desired output