Question

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

Answer 1

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

The line −2x + y = 4 meets the coordinate axis at $A\left(-2,0\right)$ and B(0, 4). Join these points to obtain the line −2x + y = 4.
Clearly, (0, 0) satisfies the inequation −2x + y ≤ 4. So, the region in xy-plane that contains the origin represents the solution set of the given equation.

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

The line x − 2y = 2 meets the coordinate axis at E(2, 0) and F(0, −1). Join these points to obtain the line x − 2y = 2.
Clearly, (0, 0) satisfies the inequation x − 2y ≤ 2. So, the region in xy-plane that 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 $B\left(0,4\right)$, $D\left(0,3\right)$ and $G\left(\frac{8}{3},\frac{1}{3}\right)$

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

Corner point	Z = 3x + 5y
$B\left(0,4\right)$	3 × 0+ 5 × 4 = 20
$D\left(0,3\right)$	3 × 0+ 5 × 3 = 15
$G\left(\frac{8}{3},\frac{1}{3}\right)$	3 × $\frac{8}{3}$+ 5 × $\frac{1}{3}$ = $\frac{29}{3}$

We see that the minimum value of the objective function Z is $\frac{29}{3}$ which is at $G\left(\frac{8}{3},\frac{1}{3}\right)$.
Thus, the optimal value of Z is $\frac{29}{3}$.