Question

The maximum value of\( z=2.5x+y\) subject to the constraints \( x+3y\leq12, 3x+y\leq12, x, y\geq0, \) is: 

• A. 4
• B. 8.5
• C. 10.5
• D. 10

Answer 1

The Correct Option is C

To find the maximum value of \( z=2.5x+y \) subject to the constraints \( x+3y\leq12, 3x+y\leq12, x\geq0, y\geq0\), we follow these steps:

1. Identify the feasible region by solving the inequalities.
2. Find the intersection points of the lines defined by \( x+3y=12 \) and \( 3x+y=12 \).

Line	Intersection with X-axis	Intersection with Y-axis
\(x+3y=12\)	\(x=12, y=0\)	\(x=0, y=4\)
\(3x+y=12\)	\(x=4, y=0\)	\(x=0, y=12\)

3. Find the intersection of \(x+3y=12\) and \(3x+y=12\):
   • Substitute \(y=12-3x\) into \(x+3(12-3x)=12\).
   • Solve: \(x+36-9x=12 \Rightarrow -8x=-24 \Rightarrow x=3, y=3\).
4. Evaluate \(z\) at intersection points:
   • \((0, 0): z=0\)
   • \((12, 0): z=30\)
   • \((4, 0): z=10\)
   • \((0, 4): z=4\)
   • \((3, 3): z=10.5\)
5. The maximum value occurs at \((3, 3)\): \(z=10.5\).

The maximum value is 10.5.