Question

solve graphically 3 X + 4 Y greater than equal to 60 x + 3 Y greater than equal to 30​

Answer 1

Answer:

3x + 4y ≤60, x + 3y≤30, x≥0, y≥0

step1:- consider the inequations as strict equations.

3x + 4y = 60

x + 3y = 30,

x = 0,

y = 0

step2:- find the points on co-ordinate axes.

for, 3x + 4y =60

when, x =0,y = 15

when, y= 0,x = 20

for, x + 3y = 30,

when, x= 0, y = 10

when,y=0, x=30

x =0 will be y-axis.

y= 0, will be x-axis.

step3:- plot the graph of ,

3x + 4y = 60,

x + 3y = 30,

x = 0,

y = 0.

step4:- take a point (0,0) and put it in inequations.

3(0)+4(0) ≤60, which is true . Hence, the shaded region will be towards the origin.

(0)+3(0)≤ 30 , which is true. Hence, the shaded region will be towards the origin.

0≥0 which is true.hence, the shaded region will be towards the origin.

0≥0 which is true.hence, the shaded region will be towards the origin.

now, see attachment.

thus, common shaded region shows the solution of the inequalities.

Answer 2

Answer:

3x + 4y ≤60, x + 3y≤30, x≥0, y≥0

step1:- consider the inequations as strict equations.

3x + 4y = 60

x + 3y = 30,

x = 0,

y = 0

step2:- find the points on co-ordinate axes.

for, 3x + 4y =60

when, x =0,y = 15

when, y= 0,x = 20

for, x + 3y = 30,

when, x= 0, y = 10

when,y=0, x=30

x =0 will be y-axis.

y= 0, will be x-axis.

step3:- plot the graph of ,

3x + 4y = 60,

x + 3y = 30,

x = 0,

y = 0.

step4:- take a point (0,0) and put it in inequations.

3(0)+4(0) ≤60, which is true . Hence, the shaded region will be towards the origin.

(0)+3(0)≤ 30 , which is true. Hence, the shaded region will be towards the origin.

0≥0 which is true.hence, the shaded region will be towards the origin.

0≥0 which is true.hence, the shaded region will be towards the origin.

Step-by-step explanation: