Consider a LPP given by
Maximize \( Z=38x+19y \)
Subjected to \( 3x+5y\le 15,\ 5x+2y\le 10 \), and \( x,y\ge 0 \)
The optimum value of \( Z \) is,
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 85
The feasible region is bounded, with corner points on the axes and at the intersection of the two lines.
\( 3x+5y=15 \) meets the axes at (5, 0) and (0, 3); \( 5x+2y=10 \) meets them at (2, 0) and (0, 5). Both constraints must hold, so the corners on the axes are (2, 0) and (0, 3).
For the intersection, multiplying the first equation by 2 and the second by 5 gives
\( 6x+10y=30 \)
\( 25x+10y=50 \)
Subtracting,
\( 19x=20 \)
\( \displaystyle x=\frac{20}{19} \)
\( \displaystyle y=\frac{15-3\cdot\frac{20}{19}}{5}=\frac{45}{19} \)
Values of Z at the corner points:
\( Z(0,0)=0 \)
\( Z(2,0)=76 \)
\( Z(0,3)=57 \)
\( \displaystyle Z\left(\frac{20}{19},\frac{45}{19}\right)=38\cdot\frac{20}{19}+19\cdot\frac{45}{19} \)
\( =40+45=85 \)
The maximum (optimum) value of Z is 85.
Why the other options are incorrect
Option (2) 190 — Incorrect. 190 is Z at (5, 0), but (5, 0) breaks \( 5x+2y\le 10 \) (25 > 10), so it is not feasible.
Option (3) 240 — Incorrect. 240 is not the value of Z at any corner point; the largest corner value is 85.
Option (4) 57 — Incorrect. 57 is Z at the corner (0, 3), which is not the largest corner value.