Respuesta :
Answer:
a) 2
b) s₁ and s₂
c) First linear equation: 5*x₁ + 8*x₂ + 10*x₃ + s₁ = 173
Second linear equation: 5*x₁ + 4*x₂ + 17*x₃ + s₂ = 254
Step-by-step explanation:
The problem statement, establishes two constraints, each one of them will need a slack variable to become a linear equation, so the answer for question
a) 2.
b) The constraints are: s₁ and s₂
c) First constraint
5*x₁ + 8*x₂ + 10*x₃ ≤ 173
We add slack variable s₁ and the inequality becomes
5*x₁ + 8*x₂ + 10*x₃ + s₁ = 173
The second constraint is:
5*x₁ + 4*x₂ + 17*x₃ ≤ 254
We add slack variable s₂ and the inequality becomes
5*x₁ + 4*x₂ + 17*x₃ + s₂ = 254
Answer:
A. Two slack variables are needed
B. S1 and s2 (option b)
C. Option d (5X1 + 8X2 + 10X3 + s1 = 173)
D. Option a (5X1 + 4X2 + 17X3 + s2 = 245)
E. Z is maximized at 173 when (X1, X2, X3) = (34.6, 0, 0)
Step-by-step explanation:
In a linear maximization problem like this, if we want to convert the inequality (constraint) into a linear equation, we add slack variables to the left hand side of each inequality.
Therefore we add s1 and s2 to the first and second inequality respectively.
[tex]5X1 + 8X2 + 10X3 \leq 173\\5X1 + 4X2 + 17X3 \leq 245[/tex]
Imputing the slack variables, we obtain as follows:
[tex]5X1 + 8X2 + 10X3 + s1 = 173\\5X1 + 4X2 + 17X3 + s2 = 245[/tex]
[tex]Maximize :\\Z = 5X1 + 3X2 + X3[/tex]
Subject to
[tex]5X1 + 8X2 + 10X3 \leq 173\\5X1 + 4X2 + 17X3 \leq 245\\With \\X1 \geq 0\\X2 \geq 0\\X3 > 0[/tex]
Solution :
[tex]5X1 + 8X2 + 10X3 + s1 = 173\\5X1 + 4X2 + 17X3 + s2 = 245\\-5X1 - 3X2 - X3 + Z = 0[/tex]
5X1 on the first equation is the pivot because the negative value of -5 is the highest and 173/5 is less than 245/5
We perform eq1 - eq2 on the first row and eq1 + eq3 on the third row to get X1 and clear the rest.
We
X1 = 173/5 = 34.6
X2 = 0 (inactive)
X3 = 0 (inactive)
s1 = 0 , s2 = -72/-1 = 72,
Z = 173/1 =173
Therefore the minimum value of Z is 173 when (X1, X2, X3) = (34.6, 0, 0)