Solve the following Linear Programming Problem graphically: Maximise Z = 5x + 3y subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.
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To solve the given LPP graphically
1) Draw all given lines,by equating to RHS
3x+5y = 15
put x= 0,y = 3
put y= 0 x = 5
Draw a line to meet the points (0,3) (5,0)
by the same way the other line
Please refer attachment for the common reason bounded by all constraints(The darkest region is the common region)
Now to,
Maximise Z = 5x + 3y
Put the coordinates of A,B,C and D
A(2,0)=> Z = 5(2) + 3(0)=10
B(1.1,2.4)=> Z= 5(1.1) + 3(2.4)=12.7
C(0,0)=> Z=5(0) + 3(0)=0
D(0,3)=> Z= 5(0) + 3(3)=9
So,the function get Maximum at x =1.1 and y = 2.4
Hope it helps you.
1) Draw all given lines,by equating to RHS
3x+5y = 15
put x= 0,y = 3
put y= 0 x = 5
Draw a line to meet the points (0,3) (5,0)
by the same way the other line
Please refer attachment for the common reason bounded by all constraints(The darkest region is the common region)
Now to,
Maximise Z = 5x + 3y
Put the coordinates of A,B,C and D
A(2,0)=> Z = 5(2) + 3(0)=10
B(1.1,2.4)=> Z= 5(1.1) + 3(2.4)=12.7
C(0,0)=> Z=5(0) + 3(0)=0
D(0,3)=> Z= 5(0) + 3(3)=9
So,the function get Maximum at x =1.1 and y = 2.4
Hope it helps you.
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