Question

Maximize Z = 4x + 3y
subject to
$$\begin{gathered} 3x+4y\le 24 \\ 8x+6y\le 48 \\ x\le 5 \\ y\le 6 \\ x,y\ge 0 \end{gathered}$$

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 × $\frac{24}{7}$ + 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.