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Answer :
- Set up the linear programming problem with the objective function and constraints.
- Use a linear programming solver to find the optimal values of the variables.
- Substitute the optimal values into the objective function to calculate the maximum value of Z.
- The maximum value of Z is $\boxed{280}$.
### Explanation
1. Problem Setup and Objective
We are given a linear programming problem with the objective to maximize $Z = 60x_1 + 30x_2 + 20x_3$ subject to the constraints:
$8x_1 + 6x_2 + x_3 \leq 48$
$4x_1 + 2x_2 + 1.5x_3 \leq 20$
$2x_1 + 1.5x_2 + 0.5x_3 \leq 8$
and $x_1, x_2, x_3 \geq 0$.
We will use the simplex method to solve this problem.
2. Applying the Simplex Method
After setting up the problem, we can use the simplex method or a linear programming solver to find the optimal solution. The result of the linear programming solver gives us the optimal values for $x_1, x_2,$ and $x_3$ that maximize the objective function $Z$.
3. Optimal Values of Variables
The linear programming solver provides the following optimal values:
$x_1 = 2$
$x_2 = 0$
$x_3 = 8$
These values satisfy all the constraints and maximize the objective function.
4. Calculating the Maximum Value of Z
Now, we substitute these optimal values into the objective function to find the maximum value of $Z$:
$Z = 60x_1 + 30x_2 + 20x_3 = 60(2) + 30(0) + 20(8) = 120 + 0 + 160 = 280$
Therefore, the maximum value of $Z$ is 280.
5. Final Answer
The maximum value of the objective function $Z = 60x_1 + 30x_2 + 20x_3$ subject to the given constraints is 280, which occurs when $x_1 = 2$, $x_2 = 0$, and $x_3 = 8$.
Thus, the solution is $\boxed{280}$.
### Examples
Linear programming is used in various real-world scenarios, such as optimizing production plans in manufacturing, determining the most efficient transportation routes for logistics, and creating investment portfolios in finance. For example, a company might use linear programming to decide how much of each product to produce in order to maximize profit, given constraints on resources like labor, materials, and machine capacity. This ensures the company operates as efficiently as possible, making the best use of available resources.
- Use a linear programming solver to find the optimal values of the variables.
- Substitute the optimal values into the objective function to calculate the maximum value of Z.
- The maximum value of Z is $\boxed{280}$.
### Explanation
1. Problem Setup and Objective
We are given a linear programming problem with the objective to maximize $Z = 60x_1 + 30x_2 + 20x_3$ subject to the constraints:
$8x_1 + 6x_2 + x_3 \leq 48$
$4x_1 + 2x_2 + 1.5x_3 \leq 20$
$2x_1 + 1.5x_2 + 0.5x_3 \leq 8$
and $x_1, x_2, x_3 \geq 0$.
We will use the simplex method to solve this problem.
2. Applying the Simplex Method
After setting up the problem, we can use the simplex method or a linear programming solver to find the optimal solution. The result of the linear programming solver gives us the optimal values for $x_1, x_2,$ and $x_3$ that maximize the objective function $Z$.
3. Optimal Values of Variables
The linear programming solver provides the following optimal values:
$x_1 = 2$
$x_2 = 0$
$x_3 = 8$
These values satisfy all the constraints and maximize the objective function.
4. Calculating the Maximum Value of Z
Now, we substitute these optimal values into the objective function to find the maximum value of $Z$:
$Z = 60x_1 + 30x_2 + 20x_3 = 60(2) + 30(0) + 20(8) = 120 + 0 + 160 = 280$
Therefore, the maximum value of $Z$ is 280.
5. Final Answer
The maximum value of the objective function $Z = 60x_1 + 30x_2 + 20x_3$ subject to the given constraints is 280, which occurs when $x_1 = 2$, $x_2 = 0$, and $x_3 = 8$.
Thus, the solution is $\boxed{280}$.
### Examples
Linear programming is used in various real-world scenarios, such as optimizing production plans in manufacturing, determining the most efficient transportation routes for logistics, and creating investment portfolios in finance. For example, a company might use linear programming to decide how much of each product to produce in order to maximize profit, given constraints on resources like labor, materials, and machine capacity. This ensures the company operates as efficiently as possible, making the best use of available resources.
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