Mike wants to fence in part of his backyard. He wants the length of the fenced-in area to be at least 20 feet long, [tex]$l \geq$[/tex] 20. He has 200 feet of fencing. The inequality that models the possible perimeter of the yard is [tex]$2 l+2 w \leq$[/tex] 200.
Which are possible dimensions for Mike's backyard? Check all that apply.
A. [tex]$w=50 ft ; l=10 ft$[/tex]
B. [tex]$w=10 ft ; l=50 ft$[/tex]
C. [tex]$w=20 ft ; l=60 ft$[/tex]
D. [tex]$w=90 ft ; l=30 ft$[/tex]
E. [tex]$w=50 ft ; l=40 ft$[/tex]
Answer (2)
Check if the length l is at least 20 feet: l ≥ 20 .
Check if the perimeter 2 l + 2 w is at most 200 feet: 2 l + 2 w ≤ 200 .
The dimensions w = 10 f t ; l = 50 f t satisfy both inequalities.
The dimensions w = 20 f t ; l = 60 f t satisfy both inequalities.
The dimensions w = 50 f t ; l = 40 f t satisfy both inequalities.
Possible dimensions: w = 10 f t ; l = 50 f t , w = 20 f t ; l = 60 f t , w = 50 f t ; l = 40 f t .
Explanation
Understanding the Problem We need to check each pair of dimensions against the given inequalities to see if they are possible for Mike's backyard. The inequalities are l ≥ 20 and 2 l + 2 w ≤ 200 .
Checking Option 1 Let's check the first option: w = 50 ft, l = 10 ft. Is l ≥ 20 ? Is 10 ≥ 20 ? No, this is false. So, this option is not possible.
Checking Option 2 Let's check the second option: w = 10 ft, l = 50 ft. Is l ≥ 20 ? Is 50 ≥ 20 ? Yes, this is true. Is 2 l + 2 w ≤ 200 ? Is 2 ( 50 ) + 2 ( 10 ) ≤ 200 ? Is 100 + 20 ≤ 200 ? Is 120 ≤ 200 ? Yes, this is true. So, this option is possible.
Checking Option 3 Let's check the third option: w = 20 ft, l = 60 ft. Is l ≥ 20 ? Is 60 ≥ 20 ? Yes, this is true. Is 2 l + 2 w ≤ 200 ? Is 2 ( 60 ) + 2 ( 20 ) ≤ 200 ? Is 120 + 40 ≤ 200 ? Is 160 ≤ 200 ? Yes, this is true. So, this option is possible.
Checking Option 4 Let's check the fourth option: w = 90 ft, l = 30 ft. Is l ≥ 20 ? Is 30 ≥ 20 ? Yes, this is true. Is 2 l + 2 w ≤ 200 ? Is 2 ( 30 ) + 2 ( 90 ) ≤ 200 ? Is 60 + 180 ≤ 200 ? Is 240 ≤ 200 ? No, this is false. So, this option is not possible.
Checking Option 5 Let's check the fifth option: w = 50 ft, l = 40 ft. Is l ≥ 20 ? Is 40 ≥ 20 ? Yes, this is true. Is 2 l + 2 w ≤ 200 ? Is 2 ( 40 ) + 2 ( 50 ) ≤ 200 ? Is 80 + 100 ≤ 200 ? Is 180 ≤ 200 ? Yes, this is true. So, this option is possible.
Final Answer The possible dimensions for Mike's backyard are: w = 10 f t ; l = 50 f t w = 20 f t ; l = 60 f t w = 50 f t ; l = 40 f t
Examples
Imagine you're planning a garden. You want the length to be at least 10 feet, and you have 60 feet of fencing. This problem helps you figure out the possible widths you can use while staying within your fencing limit and meeting your minimum length requirement. Understanding these constraints ensures you don't buy too little or too much fencing and that your garden is the size you want it to be. This is a practical application of systems of inequalities in everyday life.
The possible dimensions for Mike's backyard are: w = 10 f t ; l = 50 f t , w = 20 f t ; l = 60 f t , and w = 50 f t ; l = 40 f t . Options A and D are not possible. Therefore, the correct options are B, C, and E.
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