Answer:
See below.
Explanation:
We can solve this linear programming problem using the simplex method. We will start by converting the problem into standard form
Maximize z = 3x₁ + 5x₂ + 0s₁ + 0s₂
subject to
x₁ - 5x₂ + s₁ = 35
3x₁ - 4x₂ + s₂ = 21
x₁, x₂, s₁, s₂ ≥ 0
Next, we create the initial tableau
Basis x₁ x₂ s₁ s₂ RHS
s₁ 1 -5 1 0 35
s₂ 3 -4 0 1 21
z -3 -5 0 0 0
We can see that the initial basic variables are s₁ and s₂. We will use the simplex method to find the optimal solution.
Step 1: Choose the most negative coefficient in the bottom row as the pivot element. In this case, it is -5 in the x₂ column.
Basis x₁ x₂ s₁ s₂ RHS
s₁ 1 -5 1 0 35
s₂ 3 -4 0 1 21
z -3 -5 0 0 0
Step 2: Find the row in which the pivot element creates a positive quotient when each element in that row is divided by the pivot element. In this case, we need to find the minimum positive quotient of (35/5) and (21/4). The minimum is (21/4), so we use the second row as the pivot row.
Basis x₁ x₂ s₁ s₂ RHS
s₁ 4/5 0 1/5 1 28/5
x₂ -3/4 1 0 -1/4 -21/4
z 39/4 0 15/4 3/4 105
Step 3: Use row operations to create zeros in the x₂ column.
Basis x₁ x₂ s₁ s₂ RHS
s₁ 1 0 1/4 7/20 49/10
x₂ 0 1 3/16 -1/16 -21/16
z 0 0 39/4 21/4 525/4
The optimal solution is x₁ = 49/10, x₂ = 21/16, and z = 525/4.
Therefore, the maximum value of z is 525/4, which occurs when x₁ = 49/10 and x₂ = 21/16.