Question

Consider a LPP given by
Maximize \(Z=38x+19y\)
Subject to \(3x+5y\le 15,\ 5x+2y\le 10,\) and \(x,y\ge 0\)
The optimum value of \(Z\) is,

• A. 85
• B. 190
• C. 240
• D. 57

Hint:

Find the corner points including the intersection \((20/19,\,45/19)\), then compare Z.

Answer 1

The Correct Option is A

Step 1: Set up the feasible region.
The constraints are \(3x+5y\le 15\), \(5x+2y\le 10\), \(x\ge 0\), \(y\ge 0\). The region is bounded and lies in the first quadrant, so the maximum is at a corner point.

Step 2: Find the axis intercepts.
For \(3x+5y=15\): \((5,0)\) and \((0,3)\). For \(5x+2y=10\): \((2,0)\) and \((0,5)\). The binding limits on the axes are \(x\le 2\) (from the second line) and \(y\le 3\) (from the first line).

Step 3: Find the intersection of the two lines.
Solve \(3x+5y=15\) and \(5x+2y=10\). Multiply the first by 2 and the second by 5: \(6x+10y=30\) and \(25x+10y=50\). Subtract to get \(19x=20\).
\[ x=\frac{20}{19},\qquad y=\frac{45}{19} \]

Step 4: List the corner points.
\((0,0)\), \((2,0)\), \(\left(\frac{20}{19},\frac{45}{19}\right)\) and \((0,3)\).

Step 5: Evaluate Z at each corner.
\(Z(0,0)=0\).
\(Z(2,0)=76\).
\(Z(0,3)=57\).
\(Z\left(\frac{20}{19},\frac{45}{19}\right)=38\cdot\frac{20}{19}+19\cdot\frac{45}{19}=40+45=85\).

Step 6: Choose the largest.
The largest value is 85. Option 4 (57) is the value at \((0,3)\), a smaller vertex. Options 2 and 3 are larger than any corner value, so they cannot be reached in this region.

Final Answer:
The maximum value of \(Z\) is 85, option 1. \[ \boxed{85} \]