solve graphically 1) x-2y ≤ 3; 3x+4y ≥ 12; x ≥ 0; y ≥ 0 2) 3x+2y ≤ 12; x ≥ 1; y ≥ 2
Answers
Answer:
First we solve x−2y≤3
Lets first draw graph of
x−2y=3
Putting x=0 in (1)
0−2y=3
−2y=3
y=
−2
3
y=−1.5
Putting y=0 in (1)
x−2(0)=3
x−0=3
x=3
Refer image 1.
x
y
0
−1.5
3
0
Points to be plotted are
(0,6),(3,0)
Checking for (0,0)
Putting x=0, y=0
x−2y≤3
0−2(0)≤3
0≤3
Which is true Hence origin lies in plane x−2y≤3
So, we shade left of line
Now we solve 3x+4y≥12
Lets first draw graph of
3x+4y=12
Putting x=0 in (1)
3(0)+4y=12
0+4y=12
4y=12
y=
4
12
y=3
Putting y=0 in (1)
3x+4(0)=12
3x+0=12
3x=12
x=
3
12
x=4
Refer image 2.
x
y
0
3
4
0
Points to be plotted are
(0,3),(4,0)
checking for (0,0)
Putting x=0,y=0
3+4y≥12
3(0)+2(0)≥12
0≥12
which is false
Hence origin does not lie in plane 3x+2y>6
So, we shade rights side of line.
Also y≥1
So, for all values x
Refer image 3.
x0−14
y111
Points to be plotted are
(0,1),(−1,1),(4,1)
Also x≥0
Step-by-step explanation: