Question

Solve the following L.P.P. by graphical method Maximize: z = 3x + 5y subject to x + 4y ≤ 24, 3x + y ≤ 21 x + y ≤ 9, x ≥ 0, y ≥ 0 also find maximum value of z.

Answer 1

3x+y=21

x+y=9

Subtract

3x+y−x−y=21−9

2x=12

x=6

y=9−6=3

∴B(6,3)

For C:

x+y=9

x+4y=24

Subtract (ii) from (i) we get

x+y−x−4y=9−24

−3y=−15

y=5

x=9−5=4

∴C(4,5)

Z=3x+5y

Z at O(0,0)=3(0)+5(0)=0

Z at A(7,0)=3(7)+5(0)=21

Z at B(6,3)=3(6)+5(3)=33

Z at C(4,5)=3(4)+5(5)=37

Z at D(0,6)=3(0)+5(6)=30

Thus, Z is maximized at C(4,5) and its maximum value is 37.