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Math  /  Algebra

Question- problem5 Max⁡Z=28x1+11x2 st {1.4x1+0.6x2≤2.5x1≤1x2≥2x1,x2≥0⇒Z=48;x1=1314{x1≤0 (problem6) x1≥1 (problem7) ;x2=2\begin{array}{l} \operatorname{Max} Z=28 x_{1}+11 x_{2} \quad \text { st }\left\{\begin{array}{c} 1.4 x_{1}+0.6 x_{2} \leq 2.5 \\ x_{1} \leq 1 \\ x_{2} \geq 2 \\ x_{1}, x_{2} \geq 0 \end{array}\right. \\ \Rightarrow \mathrm{Z}=48 ; x_{1}=\frac{13}{14}\left\{\begin{array}{l} x_{1} \leq 0 \text { (problem6) } \\ x_{1} \geq 1 \text { (problem7) } ; x_{2}=2 \end{array}\right. \end{array}

Studdy Solution

STEP 1

1. We are dealing with a linear programming problem where the objective is to maximize the function Z=28x1+11x2 Z = 28x_1 + 11x_2 .
2. The constraints are linear inequalities that define a feasible region.
3. The solution involves finding values of x1 x_1 and x2 x_2 that maximize Z Z while satisfying all constraints.

STEP 2

1. Identify and list all constraints.
2. Determine the feasible region defined by the constraints.
3. Evaluate the objective function at the vertices of the feasible region.
4. Identify the vertex that provides the maximum value of the objective function.

STEP 3

Identify and list all constraints from the problem:
1. 1.4x1+0.6x2≤2.5 1.4x_1 + 0.6x_2 \leq 2.5
2. x1≤1 x_1 \leq 1
3. x2≥2 x_2 \geq 2
4. x1,x2≥0 x_1, x_2 \geq 0

STEP 4

Determine the feasible region defined by the constraints. This involves graphing the inequalities on a coordinate plane and finding the region where all conditions are satisfied simultaneously.
- The line 1.4x1+0.6x2=2.5 1.4x_1 + 0.6x_2 = 2.5 is the boundary for the first constraint. - The line x1=1 x_1 = 1 is the boundary for the second constraint. - The line x2=2 x_2 = 2 is the boundary for the third constraint. - The axes x1=0 x_1 = 0 and x2=0 x_2 = 0 are the boundaries for the non-negativity constraints.

STEP 5

Solve for the intersection points of the boundary lines to determine the vertices of the feasible region:
- Intersection of 1.4x1+0.6x2=2.5 1.4x_1 + 0.6x_2 = 2.5 and x1=1 x_1 = 1 : $ 1.4(1) + 0.6x_2 = 2.5 \implies 0.6x_2 = 1.1 \implies x_2 = \frac{11}{6} \] Vertex: \( (1, \frac{11}{6}) \)
- Intersection of 1.4x1+0.6x2=2.5 1.4x_1 + 0.6x_2 = 2.5 and x2=2 x_2 = 2 : $ 1.4x_1 + 0.6(2) = 2.5 \implies 1.4x_1 + 1.2 = 2.5 \implies 1.4x_1 = 1.3 \implies x_1 = \frac{13}{14} \] Vertex: \( (\frac{13}{14}, 2) \)
- Intersection of x1=1 x_1 = 1 and x2=2 x_2 = 2 : Vertex: (1,2) (1, 2)

STEP 6

Evaluate the objective function Z=28x1+11x2 Z = 28x_1 + 11x_2 at each vertex of the feasible region:
- At (1,116) (1, \frac{11}{6}) : $ Z = 28(1) + 11\left(\frac{11}{6}\right) = 28 + \frac{121}{6} = 28 + 20.1667 \approx 48.1667 \]
- At (1314,2) (\frac{13}{14}, 2) : $ Z = 28\left(\frac{13}{14}\right) + 11(2) = 26 + 22 = 48 \]
- At (1,2) (1, 2) : $ Z = 28(1) + 11(2) = 28 + 22 = 50 \]

STEP 7

Identify the vertex that provides the maximum value of the objective function. The maximum value of Z Z is 50 50 at the vertex (1,2) (1, 2) .
The maximum value of Z Z is 50 \boxed{50} at x1=1 x_1 = 1 and x2=2 x_2 = 2 .

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