Question

Maximize \( Z = 2x+3y \), subject to the constraints:
\( x+y \le 2 \)
\( 2x+y \le 3 \)
\( x,y \ge 0 \)

• A. 5
• B. 6
• C. 7
• D. 10

Hint:

For 2-variable LPPs, the graphical method is fastest. Quickly sketch the lines, shade the feasible region, and identify the corner points. The optimal solution (max or min) is guaranteed to be at one of these corners. Always check that your intersection points satisfy all constraints.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This is a linear programming problem (LPP). The goal is to find the maximum value of a linear objective function \(Z\) given a set of linear inequality constraints, which define a feasible region. The maximum value of \(Z\) will occur at one of the vertices (corner points) of the feasible region.

Step 2: Key Formula or Approach:
1. Graph the inequalities to identify the feasible region. 2. Determine the coordinates of the vertices of this region. 3. Evaluate the objective function \(Z = 2x+3y\) at each vertex. 4. The largest value of \(Z\) found is the maximum value.

Step 3: Detailed Explanation:
The constraints are: 1. \( x \ge 0, y \ge 0 \) (First quadrant) 2. \( x+y \le 2 \) (Region below the line \(x+y=2\)) 3. \( 2x+y \le 3 \) (Region below the line \(2x+y=3\)) Let's find the vertices of the feasible region:
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 \(2x=3 \implies x=1.5\). The point is \((1.5, 0)\). (Check other constraint: \(1.5+0 \le 2\), which is true).
Vertex 3 (y-intercept): Intersection of \(x=0\) and \(x+y=2\). This gives \(y=2\). The point is \((0,2)\). (Check other constraint: \(2(0)+2 \le 3\), which is true).
Vertex 4 (Intersection of lines): Intersection of \(x+y=2\) and \(2x+y=3\). Subtracting the first equation from the second: \((2x+y) - (x+y) = 3-2 \implies x=1\). Substitute \(x=1\) into \(x+y=2\): \(1+y=2 \implies y=1\). The point is \((1,1)\). The vertices of the feasible region are (0,0), (1.5, 0), (0,2), and (1,1). Now, 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: Final Answer:
The maximum value of Z is 6, which occurs at the point (0,2).