Math Problem Statement

maximize z=-4x-y subject to the constraints: 2x+y>=4 -2x+y<=2 x>=0 y>=0

Solution

To maximize z=4xyz = -4x - y subject to the given constraints, we can use the method of Linear Programming, which involves finding the feasible region and evaluating the objective function at the vertices of this region. Here are the steps:

1. Graphing the constraints:

We have the following constraints:

  • 2x+y42x + y \geq 4 (Rewrite as y42xy \geq 4 - 2x)
  • 2x+y2-2x + y \leq 2 (Rewrite as y2+2xy \leq 2 + 2x)
  • x0x \geq 0
  • y0y \geq 0

2. Finding the intersection points:

We need to find the points where the boundary lines of the constraints intersect. These points will be the vertices of the feasible region.

  1. Intersection of 2x+y=42x + y = 4 and 2x+y=2-2x + y = 2: 2x+y=4(Eq. 1)2x + y = 4 \quad \text{(Eq. 1)} 2x+y=2(Eq. 2)-2x + y = 2 \quad \text{(Eq. 2)}

    Subtract Eq. 2 from Eq. 1: (2x+y)(2x+y)=42(2x + y) - (-2x + y) = 4 - 2 4x=2x=124x = 2 \quad \Rightarrow x = \frac{1}{2} Substitute x=12x = \frac{1}{2} into Eq. 1: 2×12+y=41+y=4y=32 \times \frac{1}{2} + y = 4 \quad \Rightarrow 1 + y = 4 \quad \Rightarrow y = 3 So, the intersection point is (12,3)\left(\frac{1}{2}, 3\right).

  2. Intersection of 2x+y=42x + y = 4 and y=0y = 0: 2x+0=4x=22x + 0 = 4 \quad \Rightarrow x = 2 So, the intersection point is (2,0)(2, 0).

  3. Intersection of 2x+y=2-2x + y = 2 and y=0y = 0: 2x+0=2x=1-2x + 0 = 2 \quad \Rightarrow x = -1 Since x0x \geq 0, this intersection is not valid.

  4. Intersection of x=0x = 0 and 2x+y=42x + y = 4: 2(0)+y=4y=42(0) + y = 4 \quad \Rightarrow y = 4 So, the intersection point is (0,4)(0, 4).

3. Evaluating the objective function:

Now that we have the vertices of the feasible region: (12,3)\left(\frac{1}{2}, 3\right), (2,0)(2, 0), and (0,4)(0, 4), we evaluate the objective function z=4xyz = -4x - y at these points:

  1. At (12,3)\left(\frac{1}{2}, 3\right): z=4×123=23=5z = -4 \times \frac{1}{2} - 3 = -2 - 3 = -5

  2. At (2,0)(2, 0): z=4×20=8z = -4 \times 2 - 0 = -8

  3. At (0,4)(0, 4): z=4×04=4z = -4 \times 0 - 4 = -4

4. Conclusion:

The maximum value of zz occurs at the point (12,3)\left(\frac{1}{2}, 3\right) with z=5z = -5.


Would you like further details on any of the steps, or have any questions about this solution?

Related questions to explore:

  1. How would the result change if the objective function was z=4x+yz = 4x + y?
  2. What would happen if the constraints were modified to be equalities rather than inequalities?
  3. How can you apply the Simplex Method to solve this linear programming problem?
  4. How do the slack variables help in converting inequalities into equations for the Simplex Method?
  5. What other methods can be used to solve linear programming problems besides graphical methods?

Tip: Always check whether the feasible region contains any points where the constraints are violated, especially when dealing with inequalities!

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Objective Function
Inequalities
Feasible Region
Graphing

Formulas

z = -4x - y
2x + y >= 4
-2x + y <= 2
x >= 0
y >= 0

Theorems

Graphical Method for Linear Programming
Optimization at Vertices of Feasible Region

Suitable Grade Level

Grades 11-12