Question

Find the maximum and minimum value of 2x + y subject to the constraints:
x + 3y ≥ 6, x − 3y ≤ 3, 3x + 4y ≤ 24, − 3x + 2y ≤ 6, 5x + y ≥ 5, x, y ≥ 0.

Answer 1

First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 4, x + y = 3, x − 2y = 2, x = 0 and y = 0.

The line x + 3y = 6 meets the coordinate axis at $A\left(6,0\right)$ and B(0, 2). Join these points to obtain the line x + 3y = 6.
Clearly, (0, 0) does not satisfies the inequation x + 3y ≥ 6. So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.

The line x − 3y = 3 meets the coordinate axis at C(3, 0) and D(0, −1). Join these points to obtain the line x − 3y = 3.
Clearly, (0, 0) satisfies the inequation x − 3y ≤ 3. So, the region in xy-plane that contains the origin represents the solution set of the given equation.

The line 3x + 4y = 24 meets the coordinate axis at E(8, 0) and F(0, 6). Join these points to obtain the line 3x + 4y = 24.
Clearly, (0, 0) satisfies the inequation 3x + 4y ≤ 24. So, the region in xy-plane that contains the origin represents the solution set of the given equation.

The line −3x + 2y = 6 meets the coordinate axis at G(−2, 0) and H(0, 3). Join these points to obtain the line −3x + 2y = 6.
Clearly, (0, 0) satisfies the inequation −3x + 2y ≤ 6. So, the region in xy-plane that contains the origin represents the solution set of the given equation.

The line 5x + y = 5 meets the coordinate axis at $I\left(1,0\right)$ and J(0, 5). Join these points to obtain the line 5x + y = 5.
Clearly, (0, 0) does not satisfies the inequation 5x + y ≥ 5. So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.

The corner points of the feasible region are P$\left(\frac{4}{13},\frac{45}{13}\right)$, K$\left(\frac{4}{3},5\right)$, L$\left(\frac{84}{13},\frac{15}{13}\right)$, M$\left(\frac{9}{2},\frac{1}{2}\right)$, N$\left(\frac{9}{14},\frac{25}{14}\right)$

The values of Z at these corner points are as follows.

Corner point	Z = 2x + y
P$\left(\frac{4}{13},\frac{45}{13}\right)$	2 × $\frac{4}{13}$+ $\frac{45}{13}$ = $\frac{53}{13}$
K$\left(\frac{4}{3},5\right)$	2 × $\frac{4}{3}$+ 5 = $\frac{23}{3}$
L$\left(\frac{84}{13},\frac{15}{13}\right)$	2 × $\frac{84}{13}$+ $\frac{15}{13}$ = $\frac{183}{13}$
M$\left(\frac{9}{2},\frac{1}{2}\right)$	2 × $\frac{9}{2}$+ $\frac{1}{2}$ = $\frac{19}{2}$
N$\left(\frac{9}{14},\frac{25}{14}\right)$	2 × $\frac{9}{14}$+ $\frac{25}{14}$ = $\frac{43}{14}$

We see that the minimum value of the objective function Z is $\frac{43}{14}$ which is at N$\left(\frac{9}{14},\frac{25}{14}\right)$ and maximum value of the objective function is $\frac{183}{13}$ which is at L$\left(\frac{84}{13},\frac{15}{13}\right)$.