Question

The maximum value of Z=5x+3y subject to the constraints \(2x+4y \leq16,\) \(3x+y\leq9\). \(x,y\geq0\) is:

• A. 12
• B. 25
• C. 19
• D. 15

Answer 1

The Correct Option is C

To find the maximum value of \(Z=5x+3y\) subject to the constraints \(2x+4y \leq 16\), \(3x+y \leq 9\), \(x,y \geq 0\), we will use the method of Linear Programming, specifically the graphical method.

First, rewrite the inequalities:

• \(2x+4y \leq 16\) simplifies to \(x+2y \leq 8\)
• \(3x+y \leq 9\)

Plot these inequalities to find the feasible region:

1. Convert each inequality to an equation:

• \(x+2y=8\)
• \(3x+y=9\)

2. Find intercepts for \(x+2y=8\):

• For \(x=0\), \(y=4\)
• For \(y=0\), \(x=8\)

3. Find intercepts for \(3x+y=9\):

• For \(x=0\), \(y=9\)
• For \(y=0\), \(x=3\)

The vertices of the feasible region are found by solving the intersection of these lines and axes:

• Intersection of \(x+2y=8\) and \(3x+y=9\):

Substitute \(y=9-3x\) into \(x+2(9-3x)=8\):

• \(x+18-6x=8\)
• \(-5x=-10\)
• \(x=2\)
• Substitute \(x=2\) into \(y=9-3x\): \(y=3\)

The vertices are \((0,0)\), \((8,0)\), \((3,0)\), and \((2,3)\).

4. Evaluate \(Z=5x+3y\) at each vertex:

• At \((0,0)\), \(Z=5(0)+3(0)=0\)
• At \((8,0)\), \(Z=5(8)+3(0)=40\)
• At \((3,0)\), \(Z=5(3)+3(0)=15\)
• At \((2,3)\), \(Z=5(2)+3(3)=19\)

The maximum value of \(Z\) is therefore 19.

The option that matches this value is 19.