Question

Find the minimum value of the objective function \(Z=3x+5y\) under the following constraints by the graphical method: \(x+3y\ge3,\ x+y\ge2,\ x\ge0,\ y\ge0\).

Hint:

Plot both boundary lines, shade the common region above both, and evaluate Z at each corner point.

Answer 1

Step 1: Understanding the Feasible Region:
Both constraints are \(\ge\) type, so the feasible region lies above both lines \(x+3y=3\) and \(x+y=2\), in the first quadrant. This region is unbounded.

Step 2: Finding the Corner Points:
\(x+3y=3\) meets the y-axis at \((0,1)\) — rejected since it fails \(x+y\ge2\) (\(1<2\)). \(x+y=2\) meets the x-axis at \((2,0)\) — rejected since it fails \(x+3y\ge3\) (\(2<3\)). Intersection of the two lines: solving \(x+y=2\Rightarrow x=2-y\), substitute: \((2-y)+3y=3\Rightarrow2y=1\Rightarrow y=\tfrac12, x=\tfrac32\).

Step 3: Listing the actual feasible corner points:
The feasible region's corners are \((0,2)\) [where \(x+y=2\) meets the y-axis], \(\left(\tfrac32,\tfrac12\right)\) [intersection point], and \((3,0)\) [where \(x+3y=3\) meets the x-axis].

Step 4: Evaluating Z at each corner:
At \((0,2)\): \(Z=0+10=10\). At \(\left(\tfrac32,\tfrac12\right)\): \(Z=\tfrac92+\tfrac52=7\). At \((3,0)\): \(Z=9+0=9\).

Final Answer:
The smallest value is \(Z=7\) at \(\left(\tfrac32,\tfrac12\right)\); since the region is unbounded but \(Z\) only increases moving away from this vertex (positive coefficients), this is the true minimum.\[ \boxed{Z_{min}=7 \text{ at } \left(\dfrac32,\dfrac12\right)} \]