Question

please show answers clearly

Given this problem Max Z = $0.3x + $0.90y Subject to: 2x + 3.2y= 160 4x + 2.0y = 240 y = 40 x.y >=0 a) Solve for the quantities of x and y which will maximize Z. The x = They b) What is the maximum value of Z? The Z

Answer 1

Given problem is linear programming , and no method precribe to solve hence solving problem by Graphical method

Max Z =$0.30 x +$0.90y

Subject to Contraints

Subject to:
2x + 3.2y <= 160
4x + 2.0y <= 240
y <= 40

Solving above contraint equation by Graphical method

2x +3.2 y =160

Taking X =0 , hence y =160/3.2=50

Putting y =0 , x = 160/2=80

(x,y ) = (0,50)(80,0)...(.1 )

Now 4x +2y =240

Putting x =0 , y =240/2 =120

Putting y = 0 , X =240/4=60

(X,y ) =(0,120) , (60,0)....(2)

And Y = 40........(3)

Plotting above equation

See image attached

y 1 120 DO 1980 60 A ex+2y7240 40 24=160 2 > 20 40 60 80 loo 120 X Value 40 2x + 3. 2 Finding eef intersecting intersecting point e 3.24 = 160 in Putting value of y eq 2x + 3.2x40 160 2x + 128 = = 160 2x = 32 X= 16 Hence the maximum quantity of x= 16 & y = = 40 0.3 X 16+ 0.90 x 40 Maximum value of Z = = 4.8 +36 = 840.8

a) Maximum quantity of X =16 and y =40

b)Maximum value of Z =$40.8

Or We can directly find value as follow

as Given y should be less than or equal to 40

and as zmax for y is highest hence 40 unit max unit that can be taken for y .

hence puting value of Y in equation 2X +3.2 y =160. We will get x =16

and if we take contraint 2 we will get X = 40 and y =40. Contraint 1 will not satisfy.

Hence maximum quantity of X =16. And Y=40