Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

# The Objective Function Z = 4x + 3y Can Be Maximised Subjected to the Constraints 3x + 4y ≤ 24, 8x + 6y ≤ 48, X ≤ 5, Y ≤ 6; X, Y ≥ 0 (A) at Only One Point (B) at Two Points Only - Mathematics

MCQ

The objective function Z = 4x + 3y can be maximised subjected to the constraints 3x + 4y ≤ 24, 8x + 6y ≤ 48, x ≤ 5, y ≤ 6; xy ≥ 0

#### Options

•  at only one point

• at two points only

•  at an infinite number of points

•  none of these

#### Solution

at an infinite number of points
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.
GRAPH

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.

As, we know that if a LPP has two optimal solution, then there are an infinite number of optimal solutions.
Therefore, the given objective function can be subjected at an infinite number of points.

Concept: Introduction of Linear Programming
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 30 Linear programming
MCQ | Q 9 | Page 67