Q. 25.0( 1 Vote )

# Maximise Z = 3x + 4y, subject to the constraints: x + y ≤ 1, x ≥ 0, y ≥ 0

Answer :

Given:

Z=3x + 4y

It is subject to constraints

x + y ≤ 1, x ≥ 0, y ≥ 0

Now let us convert the given inequalities into equation.

We obtain the following equation

x + y ≤ 1

⇒ x + y=1

x ≥ 0

⇒ x=0

y ≥ 0

⇒ y=0

The region represented by x+y≤1:

The line x + y=1 meets the coordinate axes (0,1) and (1,0) respectively. We will join these points to obtain the line x + y=1. It is clear that (0,0) satisfies the inequation x+y≤1. So the region containing the origin represents the solution set of the inequation x+y≤1

Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations.

Plotting these equations graphically, we get

The shaded region OBC shows the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.

Corner Points are O (0, 0), B (0, 1) and C (1, 0).

Now we will substitute these values in Z at each of these corner points, we get

Hence, the maximum value of Z is 4 at the point (0, 1).

Rate this question :

Using the method of integration, find the area of the region bounded by the following lines:

5x - 2y - 10 = 0

x + y - 9 = 0

2x - 5y - 4 = 0

Mathematics - Board PapersRefer to Exercise 27. (Maximum value of Z + Minimum value of Z) is equal to

Mathematics - ExemplarMaximise and Minimise Z = 3x – 4y

subject to x – 2y ≤ 0

– 3x + y ≤ 4

x – y ≤ 6

x, y ≥ 0

Mathematics - Exemplar