Question

Maximize Z = 4x + 3y
subject to

\[3x + 4y \leq 24\]
\[8x + 6y \leq 48\]
\[ x \leq 5\]
\[ y \leq 6\]
\[ x, y \geq 0\]

Answer 1

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
\[B\left( 0, 6 \right)\]	4 × 0 + 3 × 6 = 18

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.