Question

The maximum value of the object function Z = 5x + 10 y subject to the constraints x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x ≥ 0, y ≥ 0 is
(a) 300
(b) 600
(c) 400
(d) 800

Answer 1

Answer:

600

Step-by-step explanation:

Converting the given in equations into equations

x + 2y = 120

x + y = 60

x – 2y = 0

Region represented by

The line x + 2y = 120 meets the coordinate axis at points ,A(120, 0) and B(0, 60).

x + 2y = 120

A(120,);B(0,60)

Join points A and B to obtain the line,  

clearly (0, 0) satisfies the given in equation.  

So the region containing the origin represents the solution set of the in equation x + 2y ≤ 120

Region represented by x + y ≥ 60

The line x + y = 60 meets the coordinate axis at point C(60,0) and B(0, 60). x + y = 60

Join the points C to 5 to obtain a line.  

Clearly (0,0) does not satisfy the given in equation x + y ≥ 60.  

So the region opposite to the origin represents the solution set of the in equation.

Region represented by x – 2y ≥ 0

The line x = 2y meets the coordinate axis on origin (0,0) and any other points (60, 30).

x = 2y

O(0,0);E(60,30)

Join the points (0,0) and E(10,5) to obtain a line.

Clearly (0, 0) satisfies the given in equation.  

So the region containing the origin represents the solution set of the in equation x – 2y ≥ 0.

Region represented by x ≥ 0, y ≥ 0 :  

Since every point in the first quadrant satisfies these in equation.  

So, the first quadrant is the region represented by the in equation x ≥ 0 and y ≥ 0.

The shaded region ACEF represents the common region of the above in equations.

This region is the feasible region of the given LPP.

The point of intersection of lines x + 2y = 120 and x + y = 60 is (0, 60).

The point of intersection of lines x + 2y = 120 and x – 2y = 0 is (60,30) and the point of intersection of lines x + y = 60 and x – 2y = 0 is (20, 40).

The coordinates of the vertices (comer points) of the shaded feasible region are A(120,0), C(60,0), E(40, 20) and F(60, 30).

Clearly the value of Z is minimum at C(60, 0) = 300 and the maximum value of Z= 600 at every point of line AF.

Answer 2

Given: constraints x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x ≥ 0, y ≥ 0

To Find :  The maximum value of the object function Z = 5x + 10 y

(a) 300

(b) 600

(c) 400

(d) 800

Solution:

x + y ≥ 60,

Draw x + y =  60

using points ( 0 , 60) and ( 60 , 0)

Check (0 , 0)

0 + 0 = 0 ≤ 60

Hence region is on the other side of line  where origin lies

Similarly draw other

Boundary Points of common region  

(40 , 20) , ( 60 , 0) , ( 60 , 30) and ( 120 , 0)

Z  = 5x  + 10y

x       y           Z

40     20       400

60      0          300

60      30        600

120     0          600

The maximum value of the object function Z = 5x + 10 y  is 600

correct option is  b) 600

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