Question

Subject to constraints: 2x + 4y ≤ 8, 3x + y ≤ 6, x + y ≤ 4, x, y ≥ 0; The maximum value of Z = 3x + 15y is:

• A. 28
• B. 30
• C. 40
• D. 48

Answer 1

The Correct Option is B

Solution: Constraints in inequality form The constraints are:

\[2x+4y\le8,~3x+y\le6,~x+y\le4,~x\ge0,~y\ge0\]

Converting constraints into equations For solving, we consider the boundary equations of the constraints:

\[2x+4y=8~~or~~x+2y=4\]

\[3x+y=6\]

\[x+y=4\]

We find the feasible region by solving these equations.

Finding corner points of the feasible region 1. Intersection of \(x+2y=4\) and \(3x+y=6\): Solve the equations simultaneously:

\[x+2y=4~~(1)\]

\[3x+y=6~~(2)\]

From (1), \(x=4-2y\). Substitute into (2):

\[3(4-2y)+y=6\]

\[12-6y+y=6\]

\[-5y=-6\implies y=\frac{6}{5}\]

Substituting \(y=\frac{6}{5}\) into (1):

\[x+2(\frac{6}{5})=4\]

\[x+\frac{12}{5}=4\implies x=\frac{8}{5}\]

So, one corner point is:

\[(\frac{8}{5},\frac{6}{5})\]

2. Intersection of \(x+2y=4\) and \(x+y=4\): Solve the equations:

\[x+2y=4~~(1)\]

\[x+y=4~~(2)\]

From (2), \(x = 4 - y\). Substitute into (1):

\[(4 - y) + 2y = 4\]

\[4 + y = 4 \implies y = 0\]

Substituting \(y = 0\) into (2):

\[x + 0 = 4 \implies x = 4\]

So, another corner point is:

\[(4, 0)\]

3. Intersection of \(3x + y = 6\) and \(x + y = 4\): Solve the equations:

\[3x + y = 6~~(1)\]

\[x + y = 4~~(2)\]

From (2), \(y = 4 - x\). Substitute into (1):

\[3x + (4 - x) = 6\]

\[2x + 4 = 6 \implies 2x = 2 \implies x = 1\]

Substituting \(x = 1\) into (2):

\[1 + y = 4 \implies y = 3\]

So, another corner point is:

\[(1, 3)\]

Evaluating \(Z = 3x + 15y\) at each corner point:

1. At \((\frac{8}{5}, \frac{6}{5})\):

\[Z = 3(\frac{8}{5}) + 15(\frac{6}{5}) = \frac{24}{5} + \frac{90}{5} = \frac{114}{5} = 22.8\]

2. At (4, 0):

\[Z = 3(4) + 15(0) = 12\]

3. At (1, 3):

\[Z = 3(1) + 15(3) = 3 + 45 = 30\]

Conclusion: The maximum value of Z is at (1, 3):

\[30\]