The maximum value of \( z=5x+3y \) subjected to the constraints \( 2x+3y\ge 60,2x+y\le 40;x,y\ge 0 \) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 120
The feasible region lies on or above \( 2x+3y=60 \), on or below \( 2x+y=40 \), in the first quadrant. It is the bounded triangle with corners:
\( x=0 \text{ on } 2x+3y=60:\ (0,20) \)
\( x=0 \text{ on } 2x+y=40:\ (0,40) \)
Intersection of the two lines: subtracting \( 2x+y=40 \) from \( 2x+3y=60 \) gives \( 2y=20 \), so y = 10 and x = 15:
\( (15,10) \)
Value of z = 5x + 3y at the corners:
\( (0,20):\ z=60 \)
\( (0,40):\ z=120 \)
\( (15,10):\ z=75+30=105 \)
The maximum value is 120, at (0, 40).
Why the other options are incorrect
Option (1) 105 — Incorrect. 105 is the value at the corner (15, 10), which is not the largest.
Option (3) 60 — Incorrect. 60 is the minimum value, at (0, 20).
Option (4) 90 — Incorrect. 90 is not the value of z at any corner.