Step 1: Understanding the Problem:
This is a linear programming problem (LPP) aimed at maximizing a linear objective function \(Z\) subject to linear inequality constraints, which define a feasible region. The maximum of \(Z\) is located at a vertex of the feasible region.
Step 2: Method:
1. Graph the inequalities to find the feasible region.
2. Identify the vertices of this region.
3. Calculate the objective function \(Z = 2x+3y\) at each vertex.
4. The largest value of \(Z\) is the maximum value.
Step 3: Detailed Solution:
The constraints are:
1. \( x \ge 0, y \ge 0 \) (First quadrant)
2. \( x+y \le 2 \) (Below the line \(x+y=2\))
3. \( 2x+y \le 3 \) (Below the line \(2x+y=3\))
Vertices of the feasible region are found as follows:
Vertex 1 (Origin): Intersection of \(x=0\) and \(y=0\), which is \((0,0)\).
Vertex 2 (x-intercept): Intersection of \(y=0\) and \(2x+y=3\). This gives \(x=1.5\), so the point is \((1.5, 0)\). (Check: \(1.5+0 \le 2\) is true).
Vertex 3 (y-intercept): Intersection of \(x=0\) and \(x+y=2\), resulting in \(y=2\). The point is \((0,2)\). (Check: \(2(0)+2 \le 3\) is true).
Vertex 4 (Intersection of lines): Intersection of \(x+y=2\) and \(2x+y=3\).
Subtracting the equations: \((2x+y) - (x+y) = 3-2 \implies x=1\).
Substituting \(x=1\) into \(x+y=2\): \(y=1\). The point is \((1,1)\).
The vertices are (0,0), (1.5, 0), (0,2), and (1,1).
Evaluate \(Z=2x+3y\) at each vertex:
At (0,0): \( Z = 2(0) + 3(0) = 0 \)
At (1.5, 0): \( Z = 2(1.5) + 3(0) = 3 \)
At (0,2): \( Z = 2(0) + 3(2) = 6 \)
At (1,1): \( Z = 2(1) + 3(1) = 5 \)
The values of Z are 0, 3, 5, and 6. The maximum value is 6.
Step 4: Answer:
The maximum value of Z is 6, at (0,2).