Advertisement Remove all ads

# Maximize Z = 4x + 3y Subject to 3 X + 4 Y ≤ 24 8 X + 6 Y ≤ 48 X ≤ 5 Y ≤ 6 X , Y ≥ 0 - Mathematics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Sum

Maximize Z = 4x + 3y
Subject to

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

Advertisement Remove all ads

#### 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, = 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 = 5 parallel to the Y axis.
y = 6 is the line passing through = 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 x $\frac{24}{7}$= $\frac{196}{7}$ = 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 $F\left( 5, \frac{4}{3} \right)$ $E\left( \frac{24}{7}, \frac{24}{7} \right)$  Thus, the optimal value of Z is 24.

Concept: Graphical Method of Solving Linear Programming Problems
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 30 Linear programming
Exercise 30.2 | Q 13 | Page 32

#### Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications

Forgot password?