Question

The minimum value of z=3x+6y subject to the constraints \(2x+3y≤180\), \(x+y≥60\), \(x≥3y\), \(x≥0\), \(y≥0\) is:

• A. 225
• B. 180
• C. 270
• D. 250

Answer 1

The Correct Option is B

The problem requires finding the minimum value of z=3x+6y subject to the constraints:

• 2x+3y≤180
• x+y≥60
• x≥3y
• x≥0
• y≥0

We identify the feasible region by solving these inequalities:

1. From x≥3y, express as y≤x/3.
2. From 2x+3y≤180, express as y≤(180-2x)/3.
3. From x+y≥60, express as y≥60-x.

The feasible region is bounded by these equations:

1. Intersection of x=3y and 2x+3y=180:

2(3y) + 3y = 180
9y = 180
y = 20
x = 60

2. Intersection of x=3y and x+y=60:

x = 3y
3y + y = 60
y = 15
x = 45

3. Intersection of x+y=60 and 2x+3y=180:

x+y=60
y = 60-x

2x + 3(60-x) = 180
2x + 180 - 3x = 180
-x = 0
x = 0
y = 60

For vertices (0,60), (45,15), (60,20), evaluate z=3x+6y:

• (0,60): z = 3(0)+6(60) = 360
• (45,15): z = 3(45)+6(15) = 225
• (60,20): z = 3(60)+6(20) = 180

Hence, the minimum value is 180.