A LPP \( \max Z=3x+8y \) subjected to the constraints \( 2x+y\le 10 \), \( x+3y\le 15 \), \( x,y\ge 0 \), then \( \max Z \) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 41
The feasible region is bounded by the axes and the lines \( 2x+y=10 \) and \( x+3y=15 \). Its corners:
\( \displaystyle O(0,0) \)
\( \displaystyle 2x+y=10\text{ meets the x-axis at }(5,0) \)
\( \displaystyle x+3y=15\text{ meets the y-axis at }(0,5) \)
The two lines meet where \( x=15-3y \) and \( 2(15-3y)+y=10 \), that is \( 30-5y=10 \):
\( \displaystyle y=4,\ x=3 \)
By the corner point method:
\( \displaystyle Z(0,0)=0 \)
\( \displaystyle Z(5,0)=15 \)
\( \displaystyle Z(0,5)=40 \)
\( \displaystyle Z(3,4)=9+32=41 \)
The maximum is 41, at (3, 4).