22. Many rectangles have a perimeter of 81 feet. Find the dimensions for the one with the maximum enclosed area.
1 answer:
Let
x----------> the length of the rectangle
y---------> the width of the rectangle
we know that
perimeter of rectangle=2*[x+y]
area of rectangle=x*y
P=81 ft
so
81=2*[x+y]-----------> 40.5=[x+y]
case <span>A) 20.25 feet × 20.25 feet
P=2*[20.25+20.25]-----> 81 ft----------> the perimeter is ok
A=20.25*20.25-----> 410.06 ft</span>²
case <span>B) 11 feet × 29.5 feet
</span>P=2*[11+29.5]-----> 81 ft----------> the perimeter is ok
A=11*29.5-----> 324.5 ft²
case <span>C) 9 feet × 9 feet
</span>P=2*[9+9]-----> 36 ft----------> the perimeter is not 81
A=9*9-----> 81 ft²
case <span>D) 17.5 feet × 23 feet
</span>P=2*[17.5+23]-----> 81 ft----------> the perimeter is ok
A=17.5*23-----> 402.5 ft²
the answer is
case A) 20.25 feet × 20.25 feet
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