Maximize Z = 4x + 3y
subject to
\[3x + 4y \leq 24\]
\[8x + 6y \leq 48\]
\[ x \leq 5\]
\[ y \leq 6\]
\[ x, y \geq 0\]
Solution
We need to maximize Z = 4x + 3y
First, we will convert the given inequations into equations, we obtain the following equations:
3x + 4y = 24, 8x + 6y = 48, x = 5, y = 6, x = 0 and y = 0.
The line 3x + 4y = 24 meets the coordinate axis at A(8, 0) and B(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 8x + 6y = 48 meets the coordinate axis at C(6, 0) and D(0,8). Join these points to obtain the line 8x + 6y = 48.
Clearly, (0, 0) satisfies the inequation 8x + 6y ≤ 48. So, the region in xy-plane that contains the origin represents the solution set of the given equation.
x = 5 is the line passing through x = 5 parallel to the Y axis.
y = 6 is the line passing through y = 6 parallel to the X axis.
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 O(0, 0), \[G\left( 5, 0 \right)\] , \[F\left( 5, \frac{4}{3} \right)\] , \[E\left( \frac{24}{7}, \frac{24}{7} \right)\] and \[B\left( 0, 6 \right)\] The values of Z at these corner points are as follows.
| Corner point | Z = 4x + 3y | |
| O(0, 0) | 4× 0 + 3 × 0 = 0 | |
|
\[G\left( 5, 0 \right)\]
|
4 × 5 + 3 × 0 = 20 | |
|
\[F\left( 5, \frac{4}{3} \right)\]
|
4 × 5 + 3 ×\[\frac{4}{3}\]=24
|
|
|
\[E\left( \frac{24}{7}, \frac{24}{7} \right)\]
|
4 x \[\frac{24}{7}\] +3 \[\frac{24}{7}\] = \[\frac{196}{7}\] =24 | |
|
\[B\left( 0, 6 \right)\]
|
|
We see that the maximum value of the objective function Z is 24 which is at \[F\left( 5, \frac{4}{3} \right)\] and \[E\left( \frac{24}{7}, \frac{24}{7} \right)\]
Thus, the optimal value of Z is 24.
