Question 305514: can someone help me figure out this word problem: is it possible for a rectangle with a primeter of 52 centimeters to have an area of 148.75 square centimeters? explain
Found 2 solutions by Fombitz, solver91311:
Answer by Fombitz(32388)   (Show Source):
Answer by solver91311(24713)  (Show Source):
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The largest area rectangle for a given perimeter is a square that measures one-fourth of the perimeter on each side -- see proof below. You can make the area as small as you like by selecting one of the dimensions to be as small as you like, just so long as it is larger than zero. For example, consider a 25 cm by 1 cm rectangle that would have an area of 25 square centimeters. But then a 25.999995 cm by 0.000005 cm rectangle would only have an area of 0.000129999975 square centimeters. Therefore, for a rectangle with a 52 centimeter perimeter, the range of the area function:
\ =\ lw)
is
^2\ =\ 169\text{ cm^2})
Since 148.75 is in this range, such a rectangle is possible.
Largest Rectangle for a given Perimeter
The perimeter of a rectangle is given by:
Solve for 
Substitute in
to obtain
\ =\ \frac{Pw\ -\ 2w^2}{2})
Put in standard quadratic form:
\ =\ -w^2\ +\ \frac{P}{2}w)
The graph of A is a parabola, opening down. Therefore the vertex represents a maximum value of the function. The value of the independent variable at the vertex is given by
}\ =\ \frac{P}{4})
Hence, the maximum area is obtained when the width of the rectangle is  . But if the width is , then  which means that  , and finally,  . Hence the maximum area for a given perimeter is a square with sides .
John
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