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Question
Find the minimum value of 3x + 5y subject to the constraints
− 2x + y ≤ 4, x + y ≥ 3, x − 2y ≤ 2, x, y ≥ 0.
Solution
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 \[\frac{29}{3}\] .
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