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

Answer: (B)

30

Answer 2

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):

$30$