Answer:
a.(10,23)
b.(7,30)
e.(20,20)
Step-by-step explanation:
We are given that
x=Width of garden
y=Length of garden
According to question
[tex]x\geq 5[/tex]
[tex]y\geq 20[/tex]
Perimeter of rectangle=[tex]2(x+y)[/tex]
Where x=Width of rectangle
y=Length of rectangle
Fencing used =[tex]2(x+y)[/tex]
[tex]2(x+y)\leq 80[/tex]
First we change inequality into equality
[tex]x=5[/tex]...(1)
[tex]y=20[/tex]...(2)
[tex]2(x+y)=80[/tex]
[tex]x+y=40[/tex]...(3)
Substitute x=0 in equation (3)
[tex]y=40[/tex]
Substitute y=0 in equation (3)
[tex]x=40[/tex]
Substitute x=0 and y=0 in equation (3)
[tex]2(x+y)\leq 80[/tex]
[tex]0\leq 80[/tex]
It is true. Therefore, the shaded region below the line.
[tex]0\geq 5[/tex]
[tex]0\geq 20[/tex]
These are false equation .Therefore, the shaded region above the line.
Substitute x=5 in equation (3)
[tex]5+y=40[/tex]
[tex]y=40-5=35[/tex]
Substitute y=20 in equation (3)
[tex]x+20=40[/tex]
[tex]x=40-20=20[/tex]
Intersection point of equation (1) and (3) is (5,35) and intersect point of equation (2) and (3) is (20,20).
Now,
a.(10,23)
Substitute in
[tex]2(x+y)\leq 80[/tex]
[tex]2(10+23)\leq 80[/tex]
[tex]66\geq 80[/tex]
10>5 and 23>20
it satisfied all inequality.
Hence, it is a solution.
b.(7,30)
7>5
30>20
[tex]2(7+30)=74<80[/tex]
it satisfied all inequality.
Hence, it is a solution.
c.(18,25)
18>5
25>20
[tex]2(18+25)=86>80[/tex]
It does not satisfied third inequality
Hence, it is not a solution.
d.(8,35)
8>5
35>20
[tex]2(8+35)=86>80[/tex]
It does not satisfied third inequality
Hence, it is not a solution.
e.(20,20)
[tex]2(20+20)=80[/tex]
20>5
20=20
it satisfied all inequality.
Hence, it is a solution.
