Maximize Z = 3x + 2y Subject to the constraints x + 2y ≤ 10, 3x + y ≤ 15 and x ≥ 0,y ≥ 0

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1 Answer

answered Apr 27, 2020 by PritiKumari (48.0k points)
selected Apr 27, 2020 by Ruksar03

Best answer

Converting the given in equation into the equations

x + 2y = 10

3 x + y = 15

x = 0

y = o

Region represented by x + 2y ≤ 10 :

The line x + 2y = 10, meets the coordinate axis at A(10, 0) and B(0, 5).

x + 2y = 10

x	10	0
y	0	5

A(10,0);B(0,5)

Join point A(10,0) to B(0,5) to obtain the line.

Since the origin (0,0) satisfies the in equation 0 + 2(0) = 0 ≤ 10.

So the region containing origin represents the solution set of the equation.

Region represented by 3x + y ≤ 15 :

The line 3x + y = 15 meets the coordinate axis on points C(5,0) and D(0, 15).

3x + y = 15

x	5	0
y	0	15

C(5, 0); D(0, 15)

Join point C(5, 0) and D(0, 15) to obtain the line.

Since the origin (0, 0) satisfies the given in equation 3(0) + 0 = 0 ≤ 15.

So the region containing the origin, represent the solution set of the in equation.

Region represented by x ≥ 0,y ≥ 0 :

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

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

The point of intersection of the lines x + 2y = 10 and 3x + y = 15 is E where x = 4 and y = 3.

Shaded region OCEB represents the common region of the in equations. This region is the feasible solution region of the Linear Programming Problem.

The corner points of this common region are O(0, 0), C(5, 0), E(4, 3) and B(0, 5).

The value of objective function on these points is as the following table :

From the table Z = 3x + 2y is maximum at E(4, 3). Hence, maximum value of Z = 18.