Answer:
Answer:
-2l^2+50l-300 ≥ 0
l(50 − 2l) ≥ 300
(l − 15)(l − 10) ≤ 0
Step-by-step explanation:
Explanation:
Area of the kennel: l(50 − 2l)
Minimum enclosed area required: 300 square feet
The situation is represented by the non-strict inequality l(50 − 2l) ≥ 300.
Solving the inequality, gives several equivalent statements:
50l − 2l2 ≥ 300
-2l2 + 50l − 300 ≥ 0
-2(l2 − 25l + 150) ≥ 0
-2(l − 15)(l − 10) ≥ 0
(l − 15)(l − 10) ≤ 0
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